BZOJ 3504 [Cqoi2014]危桥

题解：最大流

s连as bs

at bt连t

求最大流是否==af+bf

但是这样不对

因为as可能流到bt

所以再

s连as bt

at bs连t

再求一遍最大流

相当于让as流到bt的流流不过去

少打了一个+号WA了好几发

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=109;
const int oo=10000000;

int n;
int as,at,af;
int bs,bt,bf;
int ma[maxn][maxn];

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=z;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;
queue<int>q;
int vis[maxn];
int cur[maxn];
int d[maxn];
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((e.cap>e.flow)&&(!vis[e.to])){
                vis[e.to]=1;d[e.to]=d[x]+1;
                q.push(e.to);
            }
        }
    }
    return vis[t];
}

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;
}

void Dinicinit(){
    while(!q.empty())q.pop();
    edges.clear();
    for(int i=0;i<maxn;++i)G[i].clear();
}

char ss[maxn];
int main(){
    while(scanf("%d%d%d%d%d%d%d",&n,&as,&at,&af,&bs,&bt,&bf)!=EOF){
        ++as;++at;++bs;++bt;
        for(int i=1;i<=n;++i){
            scanf("%s",ss+1);
            for(int j=1;j<=n;++j){
                if(ss[j]=='X')ma[i][j]=0;
                if(ss[j]=='O')ma[i][j]=1;
                if(ss[j]=='N')ma[i][j]=oo;
            }
        }
        
        s=n+1;t=n+2;
        int fla=1;
        
        Dinicinit();
        for(int i=1;i<=n;++i){
            for(int j=i+1;j<=n;++j){
                if(ma[i][j])Addedge(i,j,ma[i][j]);
            }
        }
        Addedge(s,as,af);
        Addedge(s,bs,bf);
        Addedge(at,t,af);
        Addedge(bt,t,bf);
        if(Maxflow()!=af+bf)fla=0;
        
        Dinicinit();
        for(int i=1;i<=n;++i){
            for(int j=i+1;j<=n;++j){
                if(ma[i][j])Addedge(i,j,ma[i][j]);
            }
        }
        Addedge(s,as,af);
        Addedge(s,bt,bf);
        Addedge(at,t,af);
        Addedge(bs,t,bf);
        if(Maxflow()!=af+bf)fla=0;
        
        if(fla)printf("Yes
");
        else printf("No
");
    }
    return 0;
}