BZOJ 1934 [Shoi2007]Vote 善意的投票

题解:

不同集合付出代价

有朋友关系的两点之间连边

s向同意的连边,不同意的向t连边

如果本来相对的朋友(x,y)都改变

那么他们中间的边一定要被割掉,还是产生1的代价

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=1009;
const int oo=1000000000;

int n,m;

struct Edge{
    int from,to,cap,flow;
};
vector<int>G[maxn];
vector<Edge>edges;
void Addedge(int x,int y,int z){
    Edge e;
    e.from=x;e.to=y;e.cap=z;e.flow=0;
    edges.push_back(e);
    e.from=y;e.to=x;e.cap=0;e.flow=0;
    edges.push_back(e);
    int c=edges.size();
    G[x].push_back(c-2);
    G[y].push_back(c-1);
}

int s,t;
int vis[maxn];
int d[maxn];
queue<int>q;
int Bfs(){
    memset(vis,0,sizeof(vis));
    vis[s]=1;d[s]=0;q.push(s);
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int i=0;i<G[x].size();++i){
            Edge e=edges[G[x][i]];
            if((!vis[e.to])&&(e.cap>e.flow)){
                d[e.to]=d[x]+1;
                vis[e.to]=1;
                q.push(e.to);
            }
        }
    }
    return vis[t];
}

int cur[maxn];
int Dfs(int x,int a){
    if((x==t)||(a==0))return a;
    
    int nowflow=0,f=0;
    for(int i=cur[x];i<G[x].size();++i){
        cur[x]=i;
        Edge e=edges[G[x][i]];
        if((d[x]+1==d[e.to])&&((f=Dfs(e.to,min(a,e.cap-e.flow)))>0)){
            nowflow+=f;
            a-=f;
            edges[G[x][i]].flow+=f;
            edges[G[x][i]^1].flow-=f;
            if(a==0)break;
        }
    }
    return nowflow;
}

int Maxflow(){
    int flow=0;
    while(Bfs()){
        memset(cur,0,sizeof(cur));
        flow+=Dfs(s,oo);
    }
    return flow;
}

int main(){
    scanf("%d%d",&n,&m);
    s=n+1;t=s+1;
    for(int i=1;i<=n;++i){
        int x;scanf("%d",&x);
        if(x==0)Addedge(s,i,1);
        else Addedge(i,t,1);
    }
    while(m--){
        int x,y;
        scanf("%d%d",&x,&y);
        Addedge(x,y,1);
        Addedge(y,x,1);
    }
    
    cout<<Maxflow()<<endl;
    return 0;
}

 

posted @ 2018-03-20 16:09  ws_zzy  阅读(71)  评论(0编辑  收藏