2244: [SDOI2011]攔截導(dǎo)彈

Description

某國為了防御敵國的導(dǎo)彈襲擊，發(fā)展出一種導(dǎo)彈攔截系統(tǒng)。但是這種導(dǎo)彈攔截系統(tǒng)有一個(gè)缺陷：雖然它的第一發(fā)炮彈能夠到達(dá)任意的高度、并且能夠攔截任意速度的導(dǎo)彈，但是以后每一發(fā)炮彈都不能高于前一發(fā)的高度，其攔截的導(dǎo)彈的飛行速度也不能大于前一發(fā)。某天，雷達(dá)捕捉到敵國的導(dǎo)彈來襲。由于該系統(tǒng)還在試用階段，所以只有一套系統(tǒng)，因此有可能不能攔截所有的導(dǎo)彈。

在不能攔截所有的導(dǎo)彈的情況下，我們當(dāng)然要選擇使國家損失最小、也就是攔截導(dǎo)彈的數(shù)量最多的方案。但是攔截導(dǎo)彈數(shù)量的最多的方案有可能有多個(gè)，如果有多個(gè)最優(yōu)方案，那么我們會(huì)隨機(jī)選取一個(gè)作為最終的攔截導(dǎo)彈行動(dòng)藍(lán)圖。

我方間諜已經(jīng)獲取了所有敵軍導(dǎo)彈的高度和速度，你的任務(wù)是計(jì)算出在執(zhí)行上述決策時(shí)，每枚導(dǎo)彈被攔截掉的概率。

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Input

第一行包含一個(gè)正整數(shù)n，表示敵軍導(dǎo)彈數(shù)量；

下面?行按順序給出了敵軍所有導(dǎo)彈信息：

第i+1行包含2個(gè)正整數(shù)h_i和v_i，分別表示第?枚導(dǎo)彈的高度和速度。

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Output

輸出包含兩行。

第一行為一個(gè)正整數(shù)，表示最多能攔截掉的導(dǎo)彈數(shù)量；

第二行包含n個(gè)0到1之間的實(shí)數(shù)，第i個(gè)數(shù)字表示第i枚導(dǎo)彈被攔截掉的概率（你可以保留任意多位有效數(shù)字）。

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Sample Input

4

3 30

4 40

6 60

3 30

Sample Output

2

0.33333 0.33333 0.33333 1.00000

【數(shù)據(jù)規(guī)模和約定】


對(duì)于100%的數(shù)據(jù)，1≤n≤5*104， 1≤hi ，vi≤109；

均勻分布著約30%的數(shù)據(jù)，所有vi均相等。

均勻分布著約50%的數(shù)據(jù)，滿足1≤hi ，vi≤1000。

HINT

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題解：

　　因?yàn)榫鹊膯栴}wa了一天

　　設(shè)定dp[i][0/1] 表示正向和反向的LIS，f[i][0/1] 為其方案數(shù)

　　繼承是一個(gè)三維偏序，一維排序，二維CDQ，三維樹狀數(shù)組，一次解決

　　下面是兩份不同的代碼

#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,double> #define MP make_pair typedef long long LL; typedef unsigned long long ULL; const long long INF = 1e18+1LL; const double pi = acos(-1.0); const int N = 2e5+10, M = 1e3+20,inf = 2e9; struct ss{int t,x,y;/*bool operator < (const ss &r) const {return x == r.x ? y<r.y : x < r.x;}*/ }a[N],t[N];bool cmp1(ss s1,ss s2) {return s1.t < s2.t; }bool cmp2(ss s1,ss s2) {if(s1.x == s2.x) {if(s1.y == s2.y) return s1.t < s2.t;else return s1.y > s2.y;}return s1.x > s2.x; }vector<double > ans; int dp[N][2],scc = 0,tag[N]; double f[N][2]; int n; pair<int,double > C[N],C1[N]; void update(int x,pii now) {for(int i = x; i ; i -= i&-i) {if(now.first == 0)C[i] = MP(0,0);else {if(C[i].first < now.first) {C[i] = now;}else if(C[i].first == now.first) {C[i].second += now.second;}}} } pii ask(int x) {pii ret = MP(0,0);for(int i = x; i <= n; i += i&-i) {if(ret.first < C[i].first) {ret = C[i];}else if(ret.first == C[i].first){ret.second += C[i].second;}}return ret;} void cdq(int ll,int rr,int p) {if(ll == rr) {if( f[ll][p] == 0) {dp[ll][p] = 1;f[ll][p] = 1;}return ;}sort(a+ll,a+rr+1,cmp1);cdq(ll,mid,p);scc++;sort(a+ll,a+rr+1,cmp2);for(int i = ll; i <= rr; ++i) {if(a[i].t <= mid) {update(a[i].y,MP(dp[a[i].t][p],f[a[i].t][p]));}else {pii now = ask(a[i].y);now.first += 1;if(now.second == 0) continue;if(now.first > dp[a[i].t][p]) {dp[a[i].t][p] = now.first;f[a[i].t][p] = now.second;}else if(now.first == dp[a[i].t][p]){f[a[i].t][p] += now.second;}}}for(int i = ll; i <= rr; ++i) {if(a[i].t <= mid) update(a[i].y,MP(0,0));}sort(a+ll,a+rr+1,cmp1);cdq(mid+1,rr,p); } int cnt[N],san[N],cnts; int main() {scanf("%d",&n);int mxxx = 0;for(int i = 1; i <= n; ++i) {scanf("%d%d",&a[i].x,&a[i].y);san[i] = a[i].y;mxxx = max(mxxx,a[i].x);a[i].t = i;}sort(san+1,san+n+1);int SA = unique(san+1,san+n+1) - san - 1;for(int i = 1; i <= n; ++i)a[i].y = lower_bound(san+1,san+SA+1, a[i].y) - san;cdq(1,n,0);sort(a+1,a+n+1,cmp1);for(int i =1; i <= n; ++i) {a[i].y = SA - a[i].y + 1;a[i].x = mxxx - a[i].x + 1;}for(int i = 1; i <= n/2; ++i) {swap(a[i].x,a[n - i + 1].x);swap(a[i].y,a[n - i + 1].y);}cdq(1,n,1);int ans = 0;for(int i = 1; i <= n; ++i) {ans = max(ans,(int)dp[i][0]);}printf("%d\n",ans);double sum = 0.0;for(int i = 1; i <= n; ++i) {if(dp[i][0] == ans) sum += (double)f[i][0] * f[n-i+1][1];}for(int i = 1; i <= n; ++i) {if(dp[i][0] + dp[n-i+1][1] - 1 == ans) {printf("%.5lf",(double)f[i][0] * f[n-i+1][1] / sum);}else printf("%.5lf",0.0);if(i == n) printf("\n");else printf(" ");}return 0; } View Code #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int N = 1e5+10, M = 1e3+20,inf = 2e9;double f[N][2]; double dp[N][2]; int ch[N]; struct ss {int t,x,y; }a[N],t1[N],t2[N]; int cmp(ss s1,ss s2) {return s1.x > s2.x; } double C[N]; double D[N]; double first; double second; void update(int x,double firs,double secon) {for(int i = x; i < N; i += i&(-i)) {if(C[i] < firs) {C[i] = firs;D[i] = secon;}else if(C[i] == firs) {D[i] += secon;}} }void ask(int x) {first = 0.0,second = 0.0;for(int i = x; i >= 1; i -= i&(-i)) {if(C[i] > first) {first = C[i];second = D[i];}else if(C[i] == first) {second += D[i];}} } void go(int x) {for(int i = x; i < N; i += i & (-i)) {C[i] = 0,D[i] = 0;} } void CDQ(int ll,int rr,int p) {if(ll == rr) {f[ll][p] = max(f[ll][p],1.0);dp[ll][p] = max(dp[ll][p],1.0);return ;}int mid = (ll+rr)>>1;CDQ(ll,mid,p);int acnt = 0, bcnt = 0, tot = 0;for(int i = ll; i <= mid; ++i) t1[++acnt] = a[i];for(int i = mid+1; i <= rr; ++i) t2[++bcnt] = a[i];sort(t1+1,t1+acnt+1,cmp);sort(t2+1,t2+bcnt+1,cmp);int pa = 1,pb = 1;while(pb <= bcnt) {while(pa <= acnt && t1[pa].x >= t2[pb].x) {update(t1[pa].y,dp[t1[pa].t][p],f[t1[pa].t][p]);ch[++tot] = t1[pa].y;pa++;}ask(t2[pb].y);//cout<<first<<" "<<second<<endl;int id = t2[pb].t;if(first + 1 > dp[id][p]) {dp[id][p] = first+1;f[id][p] = second;}else if(first + 1 == dp[id][p])f[id][p] += second;pb++;}for(int i = 1; i <= tot; ++i) go(ch[i]);CDQ(mid+1,rr,p); } int n,san[N],SA; int main() {int mx = -1;scanf("%d",&n);for(int i = 1; i <= n; ++i) {scanf("%d%d",&a[i].x,&a[i].y);san[i] = a[i].y;a[i].t = i;mx = max(a[i].x,mx);}sort(san+1,san+n+1);SA = unique(san+1,san+n+1) - san - 1;for(int i = 1; i <= n; ++i)a[i].y = lower_bound(san+1,san+SA+1,a[i].y) - san;for(int i = 1; i <= n; ++i) {a[i].y = SA - a[i].y + 1;}CDQ(1,n,0);for(int i = 1; i <= n; ++i) {a[i].y = SA - a[i].y + 1;a[i].x = mx - a[i].x + 1;}reverse(a+1,a+n+1);for(int i = 1; i <= n; ++i) {a[i].t = i;}CDQ(1,n,1);int ans = 0;for(int i = 1; i <= n; ++i) {ans = max(ans,(int)dp[i][0]);}printf("%d\n",ans);double sum = 0.0;for(int i = 1; i <= n; ++i) {if(dp[i][0] == ans) sum += (double)f[i][0] * f[n-i+1][1];}for(int i = 1; i <= n; ++i) {if(dp[i][0] + dp[n-i+1][1] - 1 == ans) {printf("%.5lf",(double)f[i][0] * f[n-i+1][1] / sum);}else printf("%.5lf",0.0);if(i == n) printf("\n");else printf(" ");}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/zxhl/p/7243204.html

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