文章目錄

• Gorgeous Sequence
• Tyvj 1518 CPU監控
• 【清華集訓2015】V
• Naive Operations
• Picture
• Walking Race

Gorgeous Sequence

HDU5306

操作

• 區間與$x$取$\mathrm{m}\mathrm{i}\mathrm{n}$
• 查詢區間最大值
• 查詢區間和

比較暴力的線段樹維護區間

• Max : 區間最大值
• sub_max : 嚴格小于最大值的區間次大值
• cnt_max : 最大值個數
• sum : 區間和
• tag : 區間取$\mathrm{m}\mathrm{i}\mathrm{n}$的標記記錄值

考慮取$\mathrm{m}\mathrm{i}\mathrm{n}$操作走到一個區間上

• 最大值都不大于取$\mathrm{m}\mathrm{i}\mathrm{n}$的對象，說明操作沒用，return

• 如果小于最大值卻嚴格大于次大值

  此時的操作相當于區間加，打標記 sum-=(Max-x)*max_cnt;Max=x

• 否則暴力走兩邊

利用勢能分析時間復雜度

• 定義勢能函數為${\sum }_i{d}_i$，$i$是線段樹上的節點，$d_i$是這個節點對應的區間中，互不相同的元素個數
• 初始時，顯然勢能是$O(\mathrm{n}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{n})$
• 暴力對長度為$n$的區間做$\mathrm{m}\mathrm{i}\mathrm{n}$，消耗的時間是$O(n)$，勢能也會減少$O(n)$

#include <cstdio> #include <iostream> using namespace std; #define int long long #define maxn 1000005 #define lson num << 1 #define rson num << 1 | 1 struct node {int Max, sub_max, cnt_max, sum, tag; }t[maxn << 2]; int a[maxn];void pushup( int num ) {if( t[lson].Max > t[rson].Max ) {t[num].Max = t[lson].Max;t[num].cnt_max = t[lson].cnt_max;t[num].sub_max = max( t[lson].sub_max, t[rson].Max );}else if( t[lson].Max < t[rson].Max ) {t[num].Max = t[rson].Max;t[num].cnt_max = t[rson].cnt_max;t[num].sub_max = max( t[rson].sub_max, t[lson].Max );}else {t[num].Max = t[lson].Max;t[num].cnt_max = t[lson].cnt_max + t[rson].cnt_max;t[num].sub_max = max( t[lson].sub_max, t[rson].sub_max );}t[num].sum = t[lson].sum + t[rson].sum; }void build( int num, int l, int r ) {t[num].Max = t[num].sub_max = t[num].cnt_max = t[num].sum = 0, t[num].tag = -1;if( l == r ) {t[num].Max = t[num].sum = a[l];t[num].sub_max = -1;t[num].cnt_max = 1;return;}int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r );pushup( num ); }void pushdown( int num ) {if( ! ~ t[num].tag ) return;int val = t[num].tag;if( t[lson].Max > val ) {t[lson].sum -= ( t[lson].Max - val ) * t[lson].cnt_max;t[lson].Max = t[lson].tag = val;}if( t[rson].Max > val ) {t[rson].sum -= ( t[rson].Max - val ) * t[rson].cnt_max;t[rson].Max = t[rson].tag = val;}t[num].tag = -1; }void modify( int num, int l, int r, int L, int R, int val ) {if( R < l or r < L or t[num].Max <= val ) return;if( L <= l and r <= R and t[num].sub_max < val ) {t[num].sum -= ( t[num].Max - val ) * t[num].cnt_max;t[num].Max = t[num].tag = val;return;}pushdown( num );int mid = l + r >> 1;modify( lson, l, mid, L, R, val );modify( rson, mid + 1, r, L, R, val );pushup( num ); }int query_max( int num, int l, int r, int L, int R ) {if( r < L or R < l ) return -1;if( L <= l and r <= R ) return t[num].Max;pushdown( num );int mid = l + r >> 1;return max( query_max( lson, l, mid, L, R ), query_max( rson, mid + 1, r, L, R ) ); }int query_sum( int num, int l, int r, int L, int R ) {if( r < L or R < l ) return 0;if( L <= l and r <= R ) return t[num].sum;pushdown( num );int mid = l + r >> 1;return query_sum( lson, l, mid, L, R ) + query_sum( rson, mid + 1, r, L, R ); }signed main() {int n, m, T, opt, l, r, x;scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld", &n, &m );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &a[i] );build( 1, 1, n );while( m -- ) {scanf( "%lld %lld %lld", &opt, &l, &r );switch ( opt ) {case 0 : { scanf( "%lld", &x ); modify( 1, 1, n, l, r, x );break; }case 1 : { printf( "%lld\n", query_max( 1, 1, n, l, r ) ); break; }case 2 : { printf( "%lld\n", query_sum( 1, 1, n, l, r ) ); break; }}}}return 0; }

Tyvj 1518 CPU監控

BZOJ3064

操作

• 區間加
• 區間賦值
• 查詢區間最大值
• 查詢區間歷史最大值

線段樹維護

• Max : 區間最大值
• Hmax : 區間歷史最大值
• add : 區間加標記
• Hadd : 區間歷史最大加標記
• tag : 區間賦值標記
• Htag : 區間歷史最大賦值標記

假想不合并標記，每個節點上有一個隊列（按照時間先進先出），放著所有曾經推過來的標記

下推標記時，將該節點上所有的標記推到兩個兒子處，并清空隊列

對于每個節點，每次有一個區間加add標記推來時，令Max←Max+add 然后Hmax←max(Hmax,Max)

設推來的加法標記分別為$add[1...k]$，其前綴和為$S[1...k]$

則打上第$i$個標記之后，Max的值為Max+S[i]

所以，Hmax的值為${\max }_{i=1}^{k}Max+S[i]=Max+{\max }_{i=1}^{k}S[i]$

因此只需記錄${\max }_{i=1}^{k}S[i]$，就能得知該標記隊列對節點的影響

合并加法標記時簡單求和，前$i$個標記合并后恰好等于$S[i]$，那么${\max }_{i=1}^{k}S[i]$可以表述為標記的歷史最大值Hadd

此題含有賦值操作，賦值操作優先級更高，做法略有不同

賦值一次相當于前面所有的加標記和當前最大值都會被改變

所以賦值的時候，先下放所有的加標記，然后清空（包括歷史）

同時，在加標記的時候如果已經有了賦值標記，直接把加標記疊加在賦值標記上

同時記錄賦值歷史的最大值，其很有可能成為歷史最大值

#include <cstdio> #include <iostream> using namespace std; #define inf 1e18 #define maxn 100005 #define int long long #define lson num << 1 #define rson num << 1 | 1 struct node {int Max, Hmax, add, Hadd, tag, Htag; }t[maxn << 2]; int a[maxn];void build( int num, int l, int r ) {t[num].Max = t[num].Hmax = t[num].tag = t[num].Htag = -inf;t[num].add = t[num].Hadd = 0;if( l == r ) { t[num].Max = t[num].Hmax = a[l]; return; }int mid = ( l + r ) >> 1;build( lson, l, mid );build( rson, mid + 1, r );t[num].Max = t[num].Hmax = max( t[lson].Max, t[rson].Max ); }void pushup( int num ) {t[num].Max = max( t[lson].Max, t[rson].Max );t[num].Hmax = max( t[lson].Hmax, t[rson].Hmax ); }void down( int num, int add, int Hadd ) {if( t[num].Htag != -inf )t[num].Htag = max( t[num].Htag, t[num].tag + Hadd ), t[num].tag += add;elset[num].Hadd = max( t[num].Hadd, t[num].add + Hadd ), t[num].add += add; }void pushdown( int num, int l, int r ) {if( t[num].tag == -inf and ! t[num].add ) return;t[num].Hmax = max( t[num].Hmax, max( t[num].Htag, t[num].Max + t[num].Hadd ) );if( t[num].add ) {t[num].Max += t[num].add;if( l ^ r ) {down( lson, t[num].add, t[num].Hadd );down( rson, t[num].add, t[num].Hadd ); } }if( t[num].tag != -inf ) {t[num].Max = t[num].tag;if( l ^ r ) {t[lson].tag = t[rson].tag = t[num].tag;t[lson].Htag = max( t[lson].Htag, t[num].Htag );t[rson].Htag = max( t[rson].Htag, t[num].Htag );}}t[num].add = t[num].Hadd = 0;t[num].tag = t[num].Htag = -inf; }void add( int num, int l, int r, int L, int R, int val ) {pushdown( num, l, r );if( R < l or r < L ) return;if( L <= l and r <= R ) {if( t[num].tag != -inf ) t[num].tag += val, t[num].Htag = max( t[num].Htag, t[num].tag );else t[num].add += val, t[num].Hadd = max( t[num].Hadd, t[num].add );pushdown( num, l, r );return;}pushdown( num, l, r );int mid = ( l + r ) >> 1;add( lson, l, mid, L, R, val );add( rson, mid + 1, r, L, R, val );pushup( num ); }void modify( int num, int l, int r, int L, int R, int val ) {pushdown( num, l, r );if( R < l or r < L ) return;if( L <= l and r <= R ) {t[num].tag = val;t[num].Htag = max( t[num].Htag, val );pushdown( num, l, r );return;}pushdown( num, l, r );int mid = ( l + r ) >> 1;modify( lson, l, mid, L, R, val );modify( rson, mid + 1, r, L, R, val );pushup( num ); }int query_max( int num, int l, int r, int L, int R ) {if( R < l or r < L ) return -inf;pushdown( num, l, r );if( L <= l and r <= R ) return t[num].Max;int mid = ( l + r ) >> 1;return max( query_max( lson, l, mid, L, R ), query_max( rson, mid + 1, r, L, R ) ); }int queryHmax( int num, int l, int r, int L, int R ) {if( R < l or r < L ) return -inf;pushdown( num, l, r );if( L <= l and r <= R ) return t[num].Hmax;int mid = ( l + r ) >> 1;return max( queryHmax( lson, l, mid, L, R ), queryHmax( rson, mid + 1, r, L, R ) ); }signed main() {int n, m, l, r, x; char opt[5];scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &a[i] );build( 1, 1, n );scanf( "%lld", &m );while( m -- ) {scanf( "%s %lld %lld", opt, &l, &r );switch ( opt[0] ) {case 'Q' : { printf( "%lld\n", query_max( 1, 1, n, l, r ) ); break; }case 'A' : { printf( "%lld\n", queryHmax( 1, 1, n, l, r ) ); break; }case 'P' : { scanf( "%lld", &x ); add( 1, 1, n, l, r, x ); break; }case 'C' : { scanf( "%lld", &x ); modify( 1, 1, n, l, r, x );break; }}}return 0; }

【清華集訓2015】V

UOJ164

整那么高大上的物理電阻屁用沒有

操作

• 區間加
• 區間減
• 區間賦值
• 查詢單點值
• 查詢單點歷史最大值

這題很巧的轉化將多個操作合并成了一種形式

將標記改寫成(a,b)形式，標記對值的貢獻轉移形式改寫成x←max(x+a,b)

• 區間加：(val,-inf)
• 區間減：(-val,-inf)
• 區間賦值：(-inf,val)

標記的合并可以推出來，因為加法對取最大值滿足分配率

• 假設原來的標記為(a1,b1)，新來標記為(a2,b2)
• $\max (x+a_1,b_1)\leftarrow (a_2,b_2)?\max (\max (x+a_1,b_1)+a_2,b_2)?\max (\max (x+a_1+a_2,b_1+a_2),b_2)$
• 新來標記合并最后即為$x\leftarrow \max (x+a_1+a_2,\max (a_2+b_1,b_2))$

考慮當前最大值的更新

兩個標記取$\mathrm{m}\mathrm{a}\mathrm{x}$相當于兩個分段一次函數取$\mathrm{m}\mathrm{a}\mathrm{x}$

$\mathrm{m}\mathrm{a}\mathrm{x}(({\mathrm{a}}_1,{\mathrm{b}}_1),({\mathrm{a}}_2,{\mathrm{b}}_2))=(\mathrm{m}\mathrm{a}\mathrm{x}({\mathrm{a}}_1,{\mathrm{a}}_2),\mathrm{m}\mathrm{a}\mathrm{x}({\mathrm{b}}_1,{\mathrm{b}}_2))$

再考慮歷史最大值的更新

(a0,b0)表示當前，(a1,b1)表示歷史最大值，(c0,d0)表示傳遞點當前，(c1,d1)表示傳遞點歷史最大

合并(a0,b0)和(c0,d0)：$(a_0+c_0,\max (b_0+c_0,d_0))$

合并(a1,b1)和(c1,d1)，先把(c1,d1)作用在(a0,b0)上，$(a_0+c_1,\max (b_0+c_1,d_1))$

再和(a1,b1)取max：$(\max (a_1,a_0+c_1),\max (b_1,b_0+c_1,d_1))$

（a,b相當于兒子，c,d代表父親的信息）

#include <cstdio> #include <iostream> using namespace std; #define maxn 500005 #define int long long #define lson num << 1 #define rson num << 1 | 1 struct node {int a, b, maxa, maxb; }t[maxn << 2]; int w[maxn]; const int inf = 1e18;void build( int num, int l, int r ) {t[num].a = t[num].maxa = 0;t[num].b = t[num].maxb = -inf;if( l == r ) return;int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r ); }void pushdown( int num ) {if( ! t[num].a and t[num].b == -inf ) return;t[lson].maxa = max( t[lson].maxa, t[lson].a + t[num].maxa );t[lson].maxb = max( t[lson].maxb, max( t[lson].b + t[num].maxa, t[num].maxb ) );t[rson].maxa = max( t[rson].maxa, t[rson].a + t[num].maxa );t[rson].maxb = max( t[rson].maxb, max( t[rson].b + t[num].maxa, t[num].maxb ) );t[lson].a = max( t[lson].a + t[num].a, -inf );t[lson].b = max( t[lson].b + t[num].a, t[num].b );t[rson].a = max( t[rson].a + t[num].a, -inf );t[rson].b = max( t[rson].b + t[num].a, t[num].b );t[num].a = t[num].maxa = 0;t[num].b = t[num].maxb = -inf; }void modify( int num, int l, int r, int L, int R, int a, int b ) {if( r < L or R < l ) return;if( L <= l and r <= R ) {t[num].a = max( t[num].a + a, -inf );t[num].b = max( t[num].b + a, b );t[num].maxa = max( t[num].a, t[num].maxa );t[num].maxb = max( t[num].b, t[num].maxb );return;}pushdown( num );int mid = l + r >> 1;modify( lson, l, mid, L, R, a, b );modify( rson, mid + 1, r, L, R, a, b ); }int query( int num, int l, int r, int pos, int k ) {if( l == r ) {if( ! k ) return max( w[l] + t[num].a, t[num].b );else return max( w[l] + t[num].maxa, t[num].maxb );}pushdown( num );int mid = l + r >> 1;if( pos <= mid ) return query( lson, l, mid, pos, k );else return query( rson, mid + 1, r, pos, k ); }signed main() {int n, m;scanf( "%lld %lld", &n, &m );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &w[i] );build( 1, 1, n );for( int i = 1, opt, l, r, x;i <= m;i ++ ) {scanf( "%lld", &opt );switch ( opt ) {case 1 : { scanf( "%lld %lld %lld", &l, &r, &x ), modify( 1, 1, n, l, r, x, 0 ); break; }case 2 : { scanf( "%lld %lld %lld", &l, &r, &x ), modify( 1, 1, n, l, r, -x, 0 ); break; }case 3 : { scanf( "%lld %lld %lld", &l, &r, &x ), modify( 1, 1, n, l, r, -inf, x ); break; }case 4 : { scanf( "%lld", &x ), printf( "%lld\n", query( 1, 1, n, x, 0 ) ); break; }case 5 : { scanf( "%lld", &x ), printf( "%lld\n", query( 1, 1, n, x, 1 ) ); break; }}}return 0; }

Naive Operations

HDU6315

剛開始區間全為$0$，每次區間只$+1$

記錄區間的最小值，

當最小值為$0$的時候，說明多疊加了一個$b_i$的倍數

答案增加$1$，然后重新把最小值置位$b_i$ （這倆操作要走到葉子節點后再實行）

否則就打上整體$?1$的標記

#include <cstdio> #include <iostream> using namespace std; #define maxn 100005 #define int long long #define lson num << 1 #define rson num << 1 | 1 struct node {int sum, tag, Min; }t[maxn << 2]; int b[maxn];void build( int num, int l, int r ) {t[num].tag = t[num].sum = 0;if( l == r ) { t[num].Min = b[l]; return; }int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r );t[num].Min = min( t[lson].Min, t[rson].Min ); }void pushdown( int num ) {if( ! t[num].tag ) return;t[lson].tag += t[num].tag;t[rson].tag += t[num].tag;t[lson].Min -= t[num].tag;t[rson].Min -= t[num].tag;t[num].tag = 0; }void modify( int num, int l, int r, int L, int R ) {if( R < l or r < L ) return;if( L <= l and r <= R ) {t[num].Min --;if( t[num].Min ) { t[num].tag ++; return; }else if( l == r ) { t[num].sum ++, t[num].Min = b[l]; return; }}pushdown( num );int mid = l + r >> 1;modify( lson, l, mid, L, R );modify( rson, mid + 1, r, L, R );t[num].sum = t[lson].sum + t[rson].sum;t[num].Min = min( t[lson].Min, t[rson].Min ); }int query( int num, int l, int r, int L, int R ) {if( r < L or R < l ) return 0;if( L <= l and r <= R ) return t[num].sum;pushdown( num );int mid = l + r >> 1;return query( lson, l, mid, L, R ) + query( rson, mid + 1, r, L, R ); }signed main() {int n, Q, l, r; char opt[10];while( ~ scanf( "%lld %lld", &n, &Q ) ) {for( int i = 1;i <= n;i ++ ) scanf( "%lld", &b[i] );build( 1, 1, n );for( int i = 1;i <= Q;i ++ ) {scanf( "%s %d %d", opt, &l, &r );if( opt[0] == 'a' ) modify( 1, 1, n, l, r );else printf( "%lld\n", query( 1, 1, n, l, r ) );}}return 0; }

Picture

HDU1828

掃描線求矩陣周長模板題

從下往上掃描線掃，橫線長度：現在這次總區間被覆蓋的長度和上一次總區間被覆蓋的長度之差的絕對值

而豎線長度為，區間的段數*2，乘以$2$是因為一段區間左右各一條

e.g. 段數$cnt$的計算

• 若區間[0,10]被[1,2][4,5]覆蓋，則cnt=2
• 若區間[0,10]被[1,3][4,5]覆蓋，則cnt=1（兩區間剛好連在一起）
• 若區間[0,10]被[1,5][2,6]覆蓋，則cnt=1（兩區間連起來還是一段）

掃描線中的線段樹的一個節點代表一個單位長度線段

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; #define maxn 200005 #define lson num << 1 #define rson num << 1 | 1 struct node {int l, r, h, op;node(){}node( int L, int R, int H, int Op ) {l = L, r = R, h = H, op = Op;} }E[maxn]; struct tree {int l, r, len, cnt, tag, flag_l, flag_r; }t[maxn << 2];bool cmp( node x, node y ) { return x.h < y.h; }void build( int num, int l, int r ) {t[num].l = l, t[num].r = r;t[num].len = t[num].cnt = t[num].tag = t[num].flag_l = t[num].flag_r = 0;if( l == r ) return;int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r ); }void pushup( int num ) {if( t[num].tag ) {t[num].len = t[num].r - t[num].l + 1;t[num].cnt = t[num].flag_l = t[num].flag_r = 1;}else if( t[num].l == t[num].r )t[num].len = t[num].cnt = t[num].flag_l = t[num].flag_r = 0;else {t[num].len = t[lson].len + t[rson].len;t[num].cnt = t[lson].cnt + t[rson].cnt - ( t[lson].flag_r and t[rson].flag_l );t[num].flag_l = t[lson].flag_l, t[num].flag_r = t[rson].flag_r;} }void modify( int num, int l, int r, int val ) {if( t[num].r < l or r < t[num].l ) return;if( l <= t[num].l and t[num].r <= r ) {t[num].tag += val;pushup( num );return;}modify( lson, l, r, val );modify( rson, l, r, val );pushup( num ); }int Fabs( int x ) { return x < 0 ? -x : x; }int main() {int n, x1, y1, x2, y2;while( ~ scanf( "%d", &n ) ) {int L = 1e9, R = -1e9;for( int i = 1;i <= n;i ++ ) {scanf( "%d %d %d %d", &x1, &y1, &x2, &y2 );E[i] = node( x1, x2, y1, 1 );E[i + n] = node( x1, x2, y2, -1 );L = min( L, x1 );R = max( R, x2 );}n <<= 1; R --;sort( E + 1, E + n + 1, cmp );build( 1, L, R );int ans = 0, lst = 0;for( int i = 1;i <= n;i ++ ) {modify( 1, E[i].l, E[i].r - 1, E[i].op );ans += Fabs( t[1].len - lst ) + 2 * t[1].cnt * ( E[i + 1].h - E[i].h );lst = t[1].len;}printf( "%d\n", ans );}return 0; }

Walking Race

POJ3162

用兩次棧模擬$dp$求出以每個點為端點的最長路徑

用線段樹維護區間內的路徑最大值/最小值

尺取法，雙指針判斷區間$[l,r]$是否在$m$限制內

#include <stack> #include <cstdio> #include <vector> #include <iostream> using namespace std; #define maxn 1000005 #define lson num << 1 #define rson num << 1 | 1 const int inf = 1e9; int n, m; bool vis[maxn]; int maxlen[maxn]; stack < int > sta; vector < pair < int, int > > G[maxn];struct node { int fa, d, Max, id, sub_max, sub_id;void clear( int f, int dis ) { fa = f, d = dis, Max = id = sub_max = sub_id = 0; } }MS[maxn];void dfs1() {for( int i = 1;i <= n;i ++ ) vis[i] = 0;sta.push( 1 ), MS[1].clear( 0, 0 );while( ! sta.empty() ) {int u = sta.top();if( ! vis[u] )for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i].first, dis = G[u][i].second;if( v == MS[u].fa ) continue;else MS[v].clear( u, dis ), sta.push( v );}else {sta.pop();for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i].first, dis = G[u][i].second;int len = MS[v].Max + dis;if( len > MS[u].Max ) {MS[u].sub_max = MS[u].Max;MS[u].sub_id = MS[u].id;MS[u].Max = len;MS[u].id = v;}else if( len > MS[u].sub_max ) {MS[u].sub_max = len;MS[u].sub_id = v;}}}vis[u] = 1;} }void dfs2() {sta.push( 1 );while( ! sta.empty() ) {int u = sta.top(); sta.pop();for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i].first;if( v == MS[u].fa ) continue;else sta.push( v );}maxlen[u] = MS[u].Max;if( u == 1 ) continue;else {int fa = MS[u].fa, len;if( MS[fa].id == u ) len = MS[fa].sub_max + MS[u].d;else len = MS[fa].Max + MS[u].d;maxlen[u] = max( maxlen[u], len );if( len > MS[u].Max ) {MS[u].sub_max = MS[u].Max;MS[u].sub_id = MS[u].id;MS[u].Max = len;MS[u].id = fa;}else if( len > MS[u].sub_max ) {MS[u].sub_max = len;MS[u].sub_id = fa;}}} }struct tree { int Min, Max; }t[maxn << 2];void build( int num, int l, int r ) {if( l == r ) { t[num].Min = t[num].Max = maxlen[l]; return; }int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r );t[num].Max = max( t[lson].Max, t[rson].Max );t[num].Min = min( t[lson].Min, t[rson].Min ); }int query_min( int num, int l, int r, int L, int R ) {if( r < L or R < l ) return inf;if( L <= l and r <= R ) return t[num].Min;int mid = l + r >> 1;return min( query_min( lson, l, mid, L, R ), query_min( rson, mid + 1, r, L, R ) ); }int query_max( int num, int l, int r, int L, int R ) {if( r < L or R < l ) return -inf;if( L <= l and r <= R ) return t[num].Max;int mid = l + r >> 1;return max( query_max( lson, l, mid, L, R ), query_max( rson, mid + 1, r, L, R ) ); }int main() {int l, r, minn, maxx;while( ~ scanf( "%d %d", &n, &m ) ) {for( int i = 1;i <= n;i ++ ) G[i].clear();for( int i = 2, fa, d;i <= n;i ++ ) {scanf( "%d %d", &fa, &d );G[fa].push_back( make_pair( i, d ) );}dfs1(), dfs2(), build( 1, 1, n );l = r = 1, minn = maxx = maxlen[1];int ans = 0;while( l <= r and r <= n ) {if( maxx - minn <= m ) {ans = max( ans, r - l + 1 );r ++;minn = min( minn, maxlen[r] );maxx = max( maxx, maxlen[r] );}else {l ++;minn = query_min( 1, 1, n, l, r );maxx = query_max( 1, 1, n, l, r );}if( ans > n - l ) break;}printf( "%d\n", ans );}return 0; }

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