LOJ 116 有源汇有上下界最大流

思路

还是模板。
主要是有源汇之后，只有t和s不满足流量平衡，只要从t向s连一条下界是0，上界INF的边就变回无源汇的情况了
然后求出可行流流量（就是t到s的边的流量或者这条边反向边流量的相反数）后，再从s到t跑一遍最大流就能求出答案了（原来的可行流已被统计，剩下的s到t路径上能流过去的流也统计了）

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#define int long long
using namespace std;
const int MAXN = 20100;
const int INF = 0x3f3f3f3f;
struct Edge{
    int u,v,cap,flow;
};
vector<int> G[MAXN];
vector<Edge> edges;
int n,m,d[MAXN],sumx,lower[MAXN],ss,tt,vis[MAXN],dep[MAXN],cur[MAXN],s,t;
void addedge(int u,int v,int cap){
    edges.push_back((Edge){u,v,cap,0});
    edges.push_back((Edge){v,u,0,0});
    int cnt=edges.size();
    G[u].push_back(cnt-2);
    G[v].push_back(cnt-1);
}
void addedge(int u,int v,int cap,int lower){
    addedge(u,v,cap-lower);
    d[u]+=lower;
    d[v]-=lower;    
}
int dfs(int x,int a,int tx){
    if(x==tx||a==0)
        return a;
    int flow=0,f;
    for(int &i=cur[x];i<G[x].size();i++){
        Edge &e = edges[G[x][i]];
        if(dep[e.v]==dep[x]+1&&(f=dfs(e.v,min(e.cap-e.flow,a),tx))>0){
            flow+=f;
            e.flow+=f;
            edges[G[x][i]^1].flow-=f;
            a-=f;
            if(!a)
                break;
        }
    }
    return flow;
}
queue<int> q;
bool bfs(int sx,int tx){
    memset(vis,0,sizeof(vis));
    dep[sx]=0;
    q.push(sx);
    vis[sx]=1;
    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.v])&&e.cap>e.flow){
                dep[e.v]=dep[x]+1;
                vis[e.v]=true;
                q.push(e.v);
            }
        }
    }
    return vis[tx];
}
int dinic(int sx,int tx){
    int flow=0;
    while(bfs(sx,tx)){
        memset(cur,0,sizeof(cur));
        flow+=dfs(sx,INF,tx);
    }
    return flow;
}
signed main(){
    // freopen("1.in","r",stdin);
    // freopen("1.out","w",stdout);
    scanf("%lld %lld %lld %lld",&n,&m,&s,&t);
    for(int i=1;i<=m;i++){
        int a,b,c;
        scanf("%lld %lld %lld %lld",&a,&b,&lower[i],&c);
        addedge(a,b,c,lower[i]);
    }
    addedge(t,s,INF,0);
    ss=MAXN-2;
    tt=MAXN-3;
    for(int i=1;i<=n;i++){
        if(d[i]>0){
            addedge(i,tt,d[i]);
            sumx+=d[i];
        }
        else{
            addedge(ss,i,-d[i]);
        }
    }
    int max_flow=dinic(ss,tt);
    // printf("sumx=%lld maxflow=%lld
",sumx,max_flow);
    if(sumx==max_flow){
        printf("%lld
",dinic(s,t));
    }
    else{
        printf("please go home to sleep
");
    }
    return 0;
}