BZOJ 3993 SDOI2015 星际战争

做法参见JSOI 冷冻波

(话说那个题目还需要写计算几何判断是否可以打到,这个题目直接给定0,1了QAQ)

二分+最大流判定即可

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#define eps 1e-6
using namespace std;
 
const int maxn=112;
const double oo=1e12;
int n,m,S,T;
double sum,ans;
int a[maxn],b[maxn];
int map[maxn][maxn];
int h[maxn],cnt=1;
int cur[maxn];
struct edge{
    int to,next;
    double w;
}G[200010];
queue<int>Q;
int vis[maxn];
void add(int x,int y,double z){
    ++cnt;
    G[cnt].to=y;G[cnt].next=h[x];G[cnt].w=z;h[x]=cnt;
    ++cnt;
    G[cnt].to=x;G[cnt].next=h[y];G[cnt].w=0;h[y]=cnt;
}
void read(int &num){
    num=0;char ch=getchar();
    while(ch<'!')ch=getchar();
    while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
}
bool BFS(){
    memset(vis,-1,sizeof(vis));
    vis[S]=1;Q.push(S);
    while(!Q.empty()){
        int u=Q.front();Q.pop();
        for(int i=h[u];i;i=G[i].next){
            int v=G[i].to;
            if(G[i].w>0&&vis[v]==-1){
                vis[v]=vis[u]+1;
                Q.push(v);
            }
        }
    }return vis[T]!=-1;
}
double DFS(int x,double f){
    if(x==T||fabs(f)<1e-10)return f;
    double w,used=0;
    for(int i=cur[x];i;i=G[i].next){
        if(vis[G[i].to]==vis[x]+1){
            w=f-used;
            w=DFS(G[i].to,min(G[i].w,w));
            G[i].w-=w;G[i^1].w+=w;
            if(G[i].w>0)cur[x]=i;
            used+=w;if(fabs(used-f)<1e-10)return used;
        }
    }
    if(fabs(used)<1e-10)vis[x]=-1;
    return used;
}
void dinic(){
    ans=0.0;
    while(BFS()){
        for(int i=S;i<=T;++i)cur[i]=h[i];
        ans+=DFS(S,oo);
    }return;
}
bool check(double k){
    S=0;T=n+m+1;memset(h,0,sizeof(h));cnt=1;
    for(int i=1;i<=m;++i)add(S,i,k*b[i]);
    for(int i=1;i<=n;++i)add(i+m,T,a[i]);
    for(int i=1;i<=m;++i){
        for(int j=1;j<=n;++j){
            if(map[i][j])add(i,j+m,oo);
        }
    }dinic();
    //printf("%.6lf
",ans);
    return fabs(sum-ans)<1e-10;
}
int main(){
    read(n);read(m);
    for(int i=1;i<=n;++i)read(a[i]),sum+=a[i];
    for(int i=1;i<=m;++i)read(b[i]);
    for(int i=1;i<=m;++i){
        for(int j=1;j<=n;++j){
            read(map[i][j]);
        }
    }
    double L=0.0,R=5000000.0;
    while(R-L>eps){
        double mid=(L+R)*0.5;
        if(check(mid))R=mid;
        else L=mid;
    }
    printf("%.6lf
",L);
    return 0;
}

  

原文地址:https://www.cnblogs.com/joyouth/p/5392592.html