BZOJ 2244: [SDOI2011]拦截导弹 DP+CDQ分治

2244: [SDOI2011]拦截导弹

Description

某国为了防御敌国的导弹袭击，发展出一种导弹拦截系统。但是这种导弹拦截系统有一个缺陷：虽然它的第一发炮弹能够到达任意的高度、并且能够拦截任意速度的导弹，但是以后每一发炮弹都不能高于前一发的高度，其拦截的导弹的飞行速度也不能大于前一发。某天，雷达捕捉到敌国的导弹来袭。由于该系统还在试用阶段，所以只有一套系统，因此有可能不能拦截所有的导弹。

在不能拦截所有的导弹的情况下，我们当然要选择使国家损失最小、也就是拦截导弹的数量最多的方案。但是拦截导弹数量的最多的方案有可能有多个，如果有多个最优方案，那么我们会随机选取一个作为最终的拦截导弹行动蓝图。

我方间谍已经获取了所有敌军导弹的高度和速度，你的任务是计算出在执行上述决策时，每枚导弹被拦截掉的概率。

Input

第一行包含一个正整数n，表示敌军导弹数量；

下面行按顺序给出了敌军所有导弹信息：

第i+1行包含2个正整数h_i和v_i，分别表示第枚导弹的高度和速度。

Output

输出包含两行。

第一行为一个正整数，表示最多能拦截掉的导弹数量；

第二行包含n个0到1之间的实数，第i个数字表示第i枚导弹被拦截掉的概率（你可以保留任意多位有效数字）。

Sample Input

3 30

4 40

6 60

3 30

Sample Output

0.33333 0.33333 0.33333 1.00000

【数据规模和约定】

对于100%的数据，1≤n≤5*104， 1≤hi ，vi≤109；

均匀分布着约30%的数据，所有vi均相等。

均匀分布着约50%的数据，满足1≤hi ，vi≤1000。

HINT

题解：

　　因为精度的问题wa了一天

　　设定dp[i][0/1] 表示正向和反向的LIS，f[i][0/1] 为其方案数

　　继承是一个三维偏序，一维排序，二维CDQ，三维树状数组，一次解决

　　下面是两份不同的代码

#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;
}

转载于:https://www.cnblogs.com/zxhl/p/7243204.html