#网络流，二分#洛谷 3324 [SDOI2015]星际战争

题目

分析

二分答案，然后建图判断可行性

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代码

#include <cstdio>
#include <cctype>
#include <queue>
#define rr register
using namespace std;
typedef long long lll;
const int N=111; lll sum,ans;
struct node{int y; lll w; int next;}e[5300];
int dis[N],as[N],k=1,n,m,a[N],b[N],Tt,Ss;
inline signed iut(){
    rr int ans=0,f=1; rr char c=getchar();
    while (!isdigit(c)) f=(c=='-')?-f:f,c=getchar();
    while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
    return ans*f;
}
inline void add(int x,int y,lll w){
	e[++k]=(node){y,w,as[x]}; as[x]=k;
	e[++k]=(node){x,0,as[y]}; as[y]=k;
}
inline lll min(lll a,lll b){return a<b?a:b;}
inline bool bfs(int Ss){
    for (rr int i=1;i<=Tt;++i) dis[i]=0;
    rr queue<int>q; q.push(Ss); dis[Ss]=1;
    while (q.size()){
        rr int x=q.front(); q.pop();
        for (rr int i=as[x];i;i=e[i].next)
        if (e[i].w>0&&!dis[e[i].y]){
            dis[e[i].y]=dis[x]+1;
            if (e[i].y==Tt) return 1;
            q.push(e[i].y);
        }
    }
    return 0;
}
inline lll dfs(int x,lll now){
    if (x==Tt||!now) return now;
    rr lll rest=0,f;
    for (rr int i=as[x];i;i=e[i].next)
    if (e[i].w>0&&dis[e[i].y]==dis[x]+1){
        rest+=(f=dfs(e[i].y,min(now-rest,e[i].w)));
        e[i].w-=f; e[i^1].w+=f;
        if (now==rest) return rest;
    }
    if (!rest) dis[x]=0;
    return rest;
}
inline bool check(lll mid){
	for (rr int i=2;i<=k;i+=2){
		if (i<2*n+2) e[i].w=mid*b[i>>1],e[i^1].w=0;
		else if (i<2*(n+m+1)) e[i].w=a[(i>>1)-n],e[i^1].w=0;
		    else e[i].w=1e15,e[i^1].w=0;
	}
	for (ans=0;bfs(Ss);) ans+=dfs(Ss,1e15);
	return ans>=sum;
}
signed main(){
	m=iut(),n=iut(),Ss=n+m+1,Tt=Ss+1,k=1;
	for (rr int i=1;i<=m;++i) a[i]=iut()*10000,sum+=a[i];
	for (rr int i=1;i<=n;++i) b[i]=iut();
	for (rr int i=1;i<=n;++i) add(Ss,i,0);
	for (rr int i=1;i<=m;++i) add(i+n,Tt,0);
	for (rr int i=1;i<=n;++i)
	for (rr int j=1;j<=m;++j){
		rr int x=iut();
		if (x) add(i,j+n,0);
	}
	rr lll l=0,r=50000000000ll;
	while (l<r){
		rr lll mid=(l+r)>>1;
		if (check(mid)) r=mid;
		    else l=mid+1;
	}
	return !printf("%lf",l/10000.0);
}