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  • BZOJ 3993: [SDOI2015]星际战争 [二分答案 二分图]

    3993: [SDOI2015]星际战争

    题意:略


    R1D2T1考了裸二分答案+二分图最大匹配...

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    typedef long long ll;
    const int N=105, M=1e5+5, INF=1e9+5;
    const double eps=1e-12;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int n, m, s, t, a[N], b[N], g[55][55], sum;
    struct edge{int v, ne; double c, f;}e[M];
    int cnt=1, h[N];
    inline void ins(int u, int v, double c) { //printf("ins %d %d %lf
    ",u,v,c);
    	e[++cnt]=(edge){v, h[u], c, 0}; h[u]=cnt;
    	e[++cnt]=(edge){u, h[v], 0, 0}; h[v]=cnt;
    }
    int q[N], head, tail, d[N], vis[N];
    bool bfs() { //puts("bfs");
    	memset(vis, 0, sizeof(vis));
    	head=tail=1;
    	q[tail++]=s; d[s]=0; vis[s]=1;
    	while(head != tail) {
    		int u=q[head++]; //printf("u %d
    ",u);
    		for(int i=h[u];i;i=e[i].ne) 
    			if(!vis[e[i].v] && e[i].c>e[i].f) {
    				vis[e[i].v]=1; d[e[i].v]=d[u]+1;
    				q[tail++]=e[i].v;
    				if(e[i].v == t) return true;
    			}
    	}
    	return false;
    }
    int cur[N];
    double dfs(int u, double a) {
    	if(u==t || a==0) return a;
    	double flow=0, f;
    	for(int &i=cur[u];i;i=e[i].ne) 
    		if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c-e[i].f)))>0 ) {
    			flow += f;
    			e[i].f += f;
    			e[i^1].f -= f;
    			a -= f;
    			if(a==0) break;
    		}
    	if(a) d[u]=-1;
    	return flow;
    }
    double dinic() {
    	double flow=0;
    	while(bfs()) {
    		for(int i=s; i<=t; i++) cur[i]=h[i];
    		flow += dfs(s, INF); //printf("flow %lf
    ",flow);
    	}
    	//printf("flow %lf
    ", flow);
    	return flow;
    }
    
    bool check(double mid) {
    	cnt=1; memset(h, 0, sizeof(h));
    	for(int i=1; i<=m; i++) ins(s, i, b[i]*mid);
    	for(int i=1; i<=n; i++) ins(m+i, t, a[i]);
    	for(int i=1; i<=m; i++)
    		for(int j=1; j<=n; j++) if(g[i][j]) ins(i, m+j, INF);
    
    	double flow=dinic();
    	return abs(flow-sum) < eps ? 1 : 0;
    }
    double l, r;
    void solve() {
    	while(r-l > 1e-4) {
    		double mid=(l+r)/2.0;
    		if(check(mid)) r=mid;
    		else l=mid;
    	}
    	printf("%lf", l);
    }
    int main() {
    	freopen("in","r",stdin);
    	n=read(); m=read(); s=0; t=n+m+1;
    	for(int i=1; i<=n; i++) a[i]=read(), r+=a[i], sum+=a[i];
    	for(int i=1; i<=m; i++) b[i]=read();
    	for(int i=1; i<=m; i++) for(int j=1; j<=n; j++) g[i][j]=read();
    	solve();
    }
    
    
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  • 原文地址:https://www.cnblogs.com/candy99/p/6675537.html
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