[最大流]zoj 3642：Just Another Information Sharing Problem

大致题意：

     有n个人，每个人知道ai个消息，并且会向周围人最多透露bi个消息。现在问对于一个人x，他最多能知道多少消息。

 

大致思路：

    网络流，分别把消息和人抽象成点，从始点向每个消息连边，容量为1，每个消息都向知道这条消息的人连边，容量也是1.每个人都连向x，容量为他最多透露的消息数。求出最大流即可。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<30;
const int nMax=1010005;
const int mMax=3000000;

class node{
    public:
    int c,u,v,next;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax];

void addedge(int u, int v,int w){   ////dinic邻接表加边
  //  cout<<u<<" add "<<v<<" "<<w<<endl;
    edge[ne].u = u;
    edge[ne].v = v;
    edge[ne].c = w;
    edge[ne].next = head[u];
    head[u] = ne ++;
    edge[ne].u = v;
    edge[ne].v = u;
    edge[ne].c = 0;
    edge[ne].next = head[v];
    head[v] = ne ++;
}

int dinic(int s, int t){                       //  dinic
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(1){
        memset(dep, -1, sizeof(dep));
        for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
            for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
                if(edge[j].c && dep[k=edge[j].v] == -1){
                    dep[k] = dep[i] + 1;
                    ps[r ++] = k;
                    if(k == t){
                        f = r; break;
                    }
                }
        if(dep[t] == -1) break;
        memcpy(cur, head, sizeof(cur));
        i = s, top = 0;
        while(1){
            if(i == t){
                for(tr =inf, k = 0; k < top; k ++)
                    if(edge[ps[k]].c < tr)
                        tr = edge[ps[f=k]].c;
                for(k = 0; k < top; k ++){
                    edge[ps[k]].c -= tr;
                    edge[ps[k]^1].c += tr;
                }
                i = edge[ps[top=f]].u;
                res += tr;
            }
            for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){
                if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
            }
            if(cur[i]){
                ps[top ++] = cur[i];
                i = edge[cur[i]].v;
            }
            else{
                if(top == 0) break;
                dep[i] = -1;
                i = edge[ps[-- top]].u;
            }
        }
    }
    return res;
}

int maxx[500];
bool mark[1000005];
int main(){
	int n,m,i,j,a,b,c;
	while(cin>>n){
		ne=2;
		memset(mark,0,sizeof(mark));
		memset(maxx,0,sizeof(maxx));
		memset(head,0,sizeof(head));
		for(i=1;i<=n;i++){
			cin>>a>>b>>maxx[i];
			for(j=0;j<a;j++){
				cin>>b;
				addedge(200+b,i,1);
				mark[b]=1;
			}
		}
		for(i=1;i<=1000000;i++){
			if(mark[i]==1){
				addedge(0,i+200,1);
			}
		}
		cin>>a;
		for(i=1;i<=n;i++){
			if(i==a)continue;
			addedge(i,a,maxx[i]);
		}
		cout<<dinic(0,a)<<endl;
	}
	return 0;
}