UVA1658 Admiral 拆点法解决结点容量(路径不能有公共点，容量为1的时候) 最小费用最大流

Posted 2020-09-28 天道酬勤007

 

/**
题目：UVA1658 Admiral
链接：https://vjudge.net/problem/UVA-1658
题意：lrj入门经典P375
求从s到t的两条不相交（除了s和t外，没有公共点）的路径，使得权值和最小。

思路：拆点法。
除了s,t外。把其他点都拆成两个。

例如点A，拆成A和A‘。A指向A‘连一条容量为1，花费为0的边。
原来指向A的，仍然指向A点。
原来A指向其他点的，由A‘指向它们。

最小费用最大流求流量为2时候的最小费用即可。

*/
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;
const int N = 2100;
struct Edge{
    int from, to, cap, flow, cost;
    Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
};
struct MCMF{
    int n, m;
    vector<Edge> edges;
    vector<int> G[N];
    int inq[N];
    int d[N];
    int p[N];
    int a[N];

    void init(int n){
        this->n = n;
        for(int i = 0; i <= n; i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int cap,long long cost){
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int s,int t,int &flow,long long &cost){
        for(int i = 0; i <= n; i++) d[i] = INF;
        memset(inq, 0, sizeof inq);
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

        queue<int>  Q;
        Q.push(s);
        while(!Q.empty()){
            int u = Q.front(); Q.pop();
            inq[u] = 0;
            for(int i = 0; i < G[u].size(); i++){
                Edge& e = edges[G[u][i]];
                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                    d[e.to] = d[u]+e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u],e.cap-e.flow);
                    if(!inq[e.to]) {Q.push(e.to); inq[e.to] = 1;}
                }
            }
        }
        if(d[t]==INF) return false;
        flow += a[t];
        cost += (long long)d[t]*(long long)a[t];
        for(int u = t; u!=s; u = edges[p[u]].from){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
        }
        ///流量为2时的最小费用。
        if(flow==2){
            return false;
        }
        return true;
    }
    int MincostMaxflow(int s,int t,long long &cost){
        int flow = 0;
        cost = 0;
        while(BellmanFord(s,t,flow,cost));
        return flow;
    }
};
vector<int>node[N];
int main()
{
    int n, m;
    while(scanf("%d%d",&n,&m)==2)
    {
        int s = 1, t = n;
        int u, v;
        long long cost;
        MCMF mcmf;
        mcmf.init(n*2);
        for(int i = 1; i <= n; i++) node[i].clear();
        for(int i = 0; i < m; i++){
            scanf("%d%d%lld",&u,&v,&cost);
            node[u].push_back(v);
            node[u].push_back(cost);
        }
        int tot = n+1;
        for(int i = 0; i < node[1].size(); i+=2){
            mcmf.AddEdge(1,node[1][i],1,node[1][i+1]);
        }
        for(int i = 2; i < n; i++){///除了源点和汇点，其他拆点
            int from = i, to = tot++;
            mcmf.AddEdge(from,to,1,0);
            for(int j = 0; j < node[i].size(); j+=2){
                mcmf.AddEdge(to,node[i][j],1,node[i][j+1]);
            }
        }
        for(int i = 0; i < node[n].size(); i+=2){
            mcmf.AddEdge(n,node[n][i],1,node[n][i+1]);
        }
        mcmf.MincostMaxflow(s,t,cost);
        printf("%lld\n",cost);
    }
    return 0;
}