http://blog.csdn.net/libin56842/article/details/8530197

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基礎可以看上面這篇文章

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風格：

maxn是題目給的最大區間，而節點數要開4倍，確切的說……

lson和rson辨別表示結點的左孩子和右孩子。

PushUp(int rt)是把當前結點的信息更新到父節點

PushDown（int rt)是把當前結點的信息更新給孩子結點。

rt表示當前子樹的根（root），也就是當前所在的結點。

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思想：

對于每個非葉節點所標示的結點 [a,b]，其做孩子表示的區間是[a,(a+b)/2]，其右孩子表示[(a+b)/2,b].

構造：

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離散化和線段樹：

題目：x軸上有若干個線段，求線段覆蓋的總長度。

普通解法：設置坐標范圍[min,max],初始化為0，然后每一段分別染色為1，最后統計1的個數，適用于線段數目少，區間范圍小。

離散化的解法：離散化就是一一映射的關系，即將一個大坐標和小坐標進行一一映射，適用于線段數目少，區間范圍大。

例如：[10000,22000],[30300,55000],[44000,60000],[55000,60000].

第一步：排序 10000 22000 30300 44000 55000 60000

第二部：編號 1 ? ? ? ?2 ? ? ? ?3 ? ? ? ? 4 ? ? ? 5 ? ? ? ? 6

第三部：用編號來代替原數，即小數代大數 。

[10000,22000]~[1,2]

[30300,55000]~[3,5]

[44000,60000]~[4,6]

[55000,60000]~[5,6]

然后再用小數進行普通解法的步驟，最后代換回去。

線段樹的解法：線段樹通過建立線段，將原來染色O(n)的復雜度減小到 log（n），適用于線段數目多，區間范圍小的情況。

離散化的線段樹：適用于線段數目多，區間范圍大的情況。

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構造：

動態數據結構：

struct node{

?node* left;

?node* right;

……

}

靜態全局數組模擬（完全二叉樹）：

struct node{

? int left;

? int right;

……

}Tree[MAXN]

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http://www.xuebuyuan.com/1470670.html

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線段樹主要用四種用法

單點更新：

模板：

單點增減，查詢線段和

struct node {int l,r,c; }T[MAXN*4];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c; }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int l,int x) {if(T[x].l == T[x].r && T[x].l == l){T[x].c += val;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){update(val,l,(x<<1) + 1);}else{update(val,l,x<<1);}PushUp(x); }int n,m,ans;void query(int l,int r,int x) {if(T[x].l == l && T[x].r == r){ans += T[x].c;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){query(l,r,(x<<1)+1);}else if(r<=mid){query(l,r,(x<<1));}else{query(l,mid,(x<<1));query(mid+1,r,(x<<1)+1);} }

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HDU 1166

#include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <stack> #include <queue> #include <cctype> #include <vector> #include <iterator> #include <set> #include <map> #include <sstream> using namespace std;#define mem(a,b) memset(a,b,sizeof(a)) #define pf printf #define sf scanf #define spf sprintf #define pb push_back #define debug printf("!\n") #define MAXN 55555 #define MAX(a,b) a>b?a:b #define blank pf("\n") #define LL long long #define ALL(x) x.begin(),x.end() #define INS(x) inserter(x,x.begin()) #define pqueue priority_queue #define INF 0x3f3f3f3fstruct node {int l,r,c; }T[MAXN*4];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c; }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int l,int x) {if(T[x].l == T[x].r && T[x].l == l){T[x].c += val;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){update(val,l,(x<<1) + 1);}else{update(val,l,x<<1);}PushUp(x); }int n,m,ans;void query(int l,int r,int x) {if(T[x].l == l && T[x].r == r){ans += T[x].c;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){query(l,r,(x<<1)+1);}else if(r<=mid){query(l,r,(x<<1));}else{query(l,mid,(x<<1));query(mid+1,r,(x<<1)+1);} }int main() {int t,i,kase=1;char d[10];sf("%d",&t);while(t--){mem(T,0);pf("Case %d:\n",kase++);sf("%d",&n);build(1,n,1);for(i=1;i<=n;i++){int tmp;sf("%d",&tmp);update(tmp,i,1);}while (sf("%s",d) != EOF){if (d[0] == 'E') break;int x, y;sf("%d%d", &x, &y);if (d[0] == 'Q'){ans = 0;query(x,y,1);pf("%d\n",ans);}if (d[0] == 'S') update(-y,x,1);if (d[0] == 'A') update(y,x,1);}}return 0; }

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單點替換，查詢線段最高

模板：

struct node {int l,r,c; }T[MAXN*4];void PushUp(int rt) {T[rt].c = max(T[rt<<1].c,T[(rt<<1)+1].c); }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int l,int x) {if(T[x].l == T[x].r && T[x].l == l){T[x].c = val;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){update(val,l,(x<<1) + 1);}else{update(val,l,x<<1);}PushUp(x); }int n,m,ans;void query(int l,int r,int x) {if(T[x].l == l && T[x].r == r){ans = max(ans,T[x].c);return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){query(l,r,(x<<1)+1);}else if(r<=mid){query(l,r,(x<<1));}else{query(l,mid,(x<<1));query(mid+1,r,(x<<1)+1);} }

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hdu 1754

這邊要注意，輸入字符不要用%c，會導致一些難以預料的問題

#include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <stack> #include <queue> #include <cctype> #include <vector> #include <iterator> #include <set> #include <map> #include <sstream> using namespace std;#define mem(a,b) memset(a,b,sizeof(a)) #define pf printf #define sf scanf #define spf sprintf #define pb push_back #define debug printf("!\n") #define MAXN 200005 #define MAX(a,b) a>b?a:b #define blank pf("\n") #define LL long long #define ALL(x) x.begin(),x.end() #define INS(x) inserter(x,x.begin()) #define pqueue priority_queue #define INF 0x3f3f3f3fstruct node {int l,r,c; }T[MAXN*4];void PushUp(int rt) {T[rt].c = max(T[rt<<1].c,T[(rt<<1)+1].c); }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int l,int x) {if(T[x].l == T[x].r && T[x].l == l){T[x].c = val;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){update(val,l,(x<<1) + 1);}else{update(val,l,x<<1);}PushUp(x); }int n,m,ans;void query(int l,int r,int x) {if(T[x].l == l && T[x].r == r){ans = max(ans,T[x].c);return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){query(l,r,(x<<1)+1);}else if(r<=mid){query(l,r,(x<<1));}else{query(l,mid,(x<<1));query(mid+1,r,(x<<1)+1);} }int main() {int t,i,kase=1;while(sf("%d%d",&n,&m)==2){build(1,n,1);for(i=1;i<=n;i++){int tmp;sf("%d",&tmp);update(tmp,i,1);}while (m--){int x,y;char d[2];sf("%s %d %d",d,&x, &y);//pf("%s %d %d\n",d,x,y);if (d[0] == 'Q'){ans = 0;query(x,y,1);pf("%d\n",ans);}if (d[0] == 'U') update(y,x,1);}}return 0; }

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成段更新

(通常這對初學者來說是一道坎),需要用到延遲標記(或者說懶惰標記),簡單來說就是每次更新的時候不要更新到底,用延遲標記使得更新延遲到下次需要更新or詢問到的時候

http://blog.sina.com.cn/s/blog_a2dce6b30101l8bi.html

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區間替換，求總和

數組要開4倍才夠

第一種思路，標記

模板：

struct node {int l,r,c,f; }T[MAXN<<2];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c;//pf("%d %d\n",rt,T[rt].c); }void PushDown(int rt,int m) {if(T[rt].f){T[rt<<1].f = T[(rt<<1) + 1].f = T[rt].f;T[rt<<1].c = T[rt].f * (m-(m>>1));T[(rt<<1)+1].c = T[rt].f * (m>>1);T[rt].f = 0;} }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 1;T[x].f = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1);PushUp(x); }void update(int val,int L,int R,int l,int r,int x) {if(L <= l && r <= R){T[x].f = val;T[x].c = val*(r-l+1);//pf("%d %d %d\n",T[x].c,l,r);return;}PushDown(x,r-l+1);//pf("%d %d %d %d %d %d\n",val,L,R,l,r,x);int mid = (l + r)>>1;if (L <= mid){update(val,L,R,l,mid,x<<1);}if(R > mid){update(val,L,R,mid+1,r,x<<1|1);}PushUp(x); }

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第二種思路，雜色

struct node {int l,r,c; }T[MAXN<<2];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c; }void PushDown(int rt) {if(T[rt].c != -1)//如果該區間只有一種顏色 {T[rt<<1].c = T[rt<<1|1].c = T[rt].c;//由于后面必定對子樹操作，所以更新子樹的值等于父親的值T[rt].c = -1;//由于該區域顏色與修改不同，而且不是給定區域，所以該區域必定為雜色 } }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 1;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int L,int R,int x) {if(T[x].c == val) return;//相同則不用修改了if(T[x].l == L && T[x].r == R)//找到了區間，直接更新 {T[x].c = val;return;}PushDown(x);//父區間為雜色時對所有子節點進行操作int mid = (T[x].l + T[x].r)>>1;if(L>mid)update(val,L,R,x<<1|1);else if(R<=mid)update(val,L,R,x<<1);else{update(val,L,mid,x<<1);update(val,mid+1,R,x<<1|1);} }

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hdu 1698

http://www.tuicool.com/articles/j6N3eaz

這里鏈接的其實不對，要求總和，所以每個點不能初始化為1

#include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <stack> #include <queue> #include <cctype> #include <vector> #include <iterator> #include <set> #include <map> #include <sstream> using namespace std;#define mem(a,b) memset(a,b,sizeof(a)) #define pf printf #define sf scanf #define spf sprintf #define pb push_back #define debug printf("!\n") #define MAXN 100000 + 5 #define MAX(a,b) a>b?a:b #define blank pf("\n") #define LL long long #define ALL(x) x.begin(),x.end() #define INS(x) inserter(x,x.begin()) #define pqueue priority_queue #define INF 0x3f3f3f3fstruct node {int l,r,c,f; }T[MAXN<<2];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c;//pf("%d %d\n",rt,T[rt].c); }void PushDown(int rt,int m) {if(T[rt].f){T[rt<<1].f = T[(rt<<1) + 1].f = T[rt].f;T[rt<<1].c = T[rt].f * (m-(m>>1));T[(rt<<1)+1].c = T[rt].f * (m>>1);T[rt].f = 0;} }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].c = 1;T[x].f = 0;if (l == r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1);PushUp(x); }void update(int val,int L,int R,int l,int r,int x) {if(L <= l && r <= R){T[x].f = val;T[x].c = val*(r-l+1);//pf("%d %d %d\n",T[x].c,l,r);return;}PushDown(x,r-l+1);//pf("%d %d %d %d %d %d\n",val,L,R,l,r,x);int mid = (l + r)>>1;if (L <= mid){update(val,L,R,l,mid,x<<1);}if(R > mid){update(val,L,R,mid+1,r,x<<1|1);}PushUp(x); }int n,m,ans;void query(int l,int r,int x) {if(T[x].l == l && T[x].r == r){ans += T[x].c;return;}int mid = (T[x].l + T[x].r)>>1;if (l > mid){query(l,r,(x<<1)+1);}else if(r<=mid){query(l,r,(x<<1));}else{query(l,mid,(x<<1));query(mid+1,r,(x<<1)+1);} }int a[MAXN];int main() {int t,i,kase=1;sf("%d",&t);while(t--){sf("%d",&n);build(1,n,1);sf("%d",&m);for(i=1;i<=m;i++){int x,y,z;sf("%d%d%d",&x,&y,&z);update(z,x,y,1,n,1);}pf("Case %d: The total value of the hook is %d.\n",kase++,T[1].c);}return 0; }

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區間增減，區間求和

模板：

struct node {LL l,r,c,f; }T[MAXN<<2];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c;//pf("%d %d\n",rt,T[rt].c); }void PushDown(int rt,int m) {if(T[rt].f){T[rt<<1].f += T[rt].f;T[(rt<<1) + 1].f += T[rt].f;T[rt<<1].c += T[rt].f * (m-(m>>1));T[(rt<<1)+1].c += T[rt].f * (m>>1);T[rt].f = 0;} }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].f = 0;T[x].c = 0;if(l==r) return;int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1); }void update(int val,int L,int R,int l,int r,int x) {if(L <= l && r <= R){T[x].f += val;T[x].c += val*(r-l+1);//pf("%d %d %d\n",T[x].c,l,r);return;}PushDown(x,r-l+1);//pf("%d %d %d %d %d %d\n",val,L,R,l,r,x);int mid = (l + r)>>1;if (L <= mid){update(val,L,R,l,mid,x<<1);}if(R > mid){update(val,L,R,mid+1,r,x<<1|1);}PushUp(x); }LL ans;int n,m;void query(int L,int R,int l,int r,int x) {if(L <= l && r <= R){ans += T[x].c;return;}PushDown(x,r-l+1);int mid = (l + r)>>1;if(L <= mid) query(L,R,l,mid,x<<1);if(R > mid) query(L,R,mid+1,r,x<<1|1);PushUp(x); }

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poj 3468

#include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <stack> #include <queue> #include <cctype> #include <vector> #include <iterator> #include <set> #include <map> #include <sstream> using namespace std;#define mem(a,b) memset(a,b,sizeof(a)) #define pf printf #define sf scanf #define spf sprintf #define pb push_back #define debug printf("!\n") #define MAXN 111111 + 5 #define MAX(a,b) a>b?a:b #define blank pf("\n") #define LL long long #define ALL(x) x.begin(),x.end() #define INS(x) inserter(x,x.begin()) #define pqueue priority_queue #define INF 0x3f3f3f3fstruct node {LL l,r,c,f; }T[MAXN<<2];void PushUp(int rt) {T[rt].c = T[rt<<1].c + T[(rt<<1)+1].c;//pf("%d %d\n",rt,T[rt].c); }void PushDown(int rt,int m) {if(T[rt].f){T[rt<<1].f += T[rt].f;T[(rt<<1) + 1].f += T[rt].f;T[rt<<1].c += T[rt].f * (m-(m>>1));T[(rt<<1)+1].c += T[rt].f * (m>>1);T[rt].f = 0;} }void build(int l,int r,int x) {T[x].l = l;T[x].r = r;T[x].f = 0;if(l==r){scanf("%I64d",&T[x].c);return;}int mid = (l+r)>>1;build(l,mid,x<<1);build(mid+1,r,(x<<1) + 1);PushUp(x); }void update(int val,int L,int R,int l,int r,int x) {if(L <= l && r <= R){T[x].f += val;T[x].c += val*(r-l+1);//pf("%d %d %d\n",T[x].c,l,r);return;}PushDown(x,r-l+1);//pf("%d %d %d %d %d %d\n",val,L,R,l,r,x);int mid = (l + r)>>1;if (L <= mid){update(val,L,R,l,mid,x<<1);}if(R > mid){update(val,L,R,mid+1,r,x<<1|1);}PushUp(x); }LL ans;int n,m;void query(int L,int R,int l,int r,int x) {if(L <= l && r <= R){ans += T[x].c;return;}PushDown(x,r-l+1);int mid = (l + r)>>1;if(L <= mid) query(L,R,l,mid,x<<1);if(R > mid) query(L,R,mid+1,r,x<<1|1); }int main() {int t,i,kase=1;while(~sf("%d%d",&n,&m)){build(1,n,1);/*for(i=1;i<=n;i++){int tmp;sf("%d",&tmp);update(tmp,i,i,1,n,1);}*/for(i=1;i<=m;i++){int x,y,z;char s[2];sf("%s",s);if(s[0]=='Q'){ans = 0;sf("%d%d",&x,&y);query(x,y,1,n,1);pf("%I64d\n",ans);}else{sf("%d%d%d",&x,&y,&z);update(z,x,y,1,n,1);}}}return 0; }

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轉載于:https://www.cnblogs.com/qlky/p/5690265.html

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