LOJ 115 无源汇有上下界可行流

思路

模板题
首先把下界去掉（连边upper-lower
然后记录每个点至少要流出的和流入的流量（下界
然后流入大于流出的连向tt，流出大于流入的连向ss（补流
然后跑最大流，如果sum=flow，就代表加上的边满流，有答案，否则没有

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],id[50000];
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){
    if(x==tt||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)))>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(void){
    memset(vis,0,sizeof(vis));
    dep[ss]=0;
    q.push(ss);
    vis[ss]=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[tt];
}
int dinic(void){
    int flow=0;
    while(bfs()){
        memset(cur,0,sizeof(cur));
        flow+=dfs(ss,INF);
    }
    return flow;
}
signed main(){
    // freopen("1.in","r",stdin);
    // freopen("1.out","w",stdout);
    scanf("%lld %lld",&n,&m);
    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]);
        id[i]=edges.size()-2;
    }
    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();
    // printf("sumx=%lld maxflow=%lld
",sumx,max_flow);
    if(sumx==max_flow){
        printf("YES
");
        for(int i=1;i<=m;i++)
            printf("%lld
",lower[i]+edges[id[i]].flow);
    }
    else{
        printf("NO
");
    }
    return 0;
}