• LOJ 117 有源汇有上下界最小流


    思路

    首先加上t-s的边跑最大流
    设d是加上的边的流量,删掉加上的边
    答案是d-maxflow(t,s)

    代码

    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <queue>
    #include <vector>
    #define int long long
    using namespace std;
    const int MAXN = 50100;
    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[200000],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]);
        }
        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]);
            }
        }
        addedge(t,s,INF,0);
        int max_flow=dinic(ss,tt);
        // printf("sumx=%lld maxflow=%lld
    ",sumx,max_flow);
        if(sumx==max_flow){
            int d=edges[G[t][G[t].size()-1]].flow;
            G[s].pop_back();
            G[t].pop_back();
            printf("%lld
    ",d-dinic(t,s));
        }
        else{
            printf("please go home to sleep
    ");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dreagonm/p/10803040.html
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