UVa #1658 Admiral (例题11-9)

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每个点只能访问一次，可以将每个点拆成两个点，中间连一条费用为0、容量为1的边。这样求最小费用最大流时就保证了每个点只访问一次。

因为是两艘船走，所以限制流量为2，求出最小费用即可。注意点数会翻倍，数组也要多开一倍大。

Run Time: 0.046s

#define UVa  "LT11-9.1658.cpp"          //Admiral#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<iostream>using namespace std;//Global Variables.const int maxn = 2000 + 5, INF = 1<<30;int v, e;////struct Edge {    int from, to, cap, flow, weight;    Edge(int a, int b, int c, int d): from(a), to(b), cap(c), weight(d), flow(0) {}};struct MCMF {    vector<Edge> edges;    vector<int> G[maxn];    int a[maxn], p[maxn], inq[maxn], d[maxn];    void addEdge(int from, int to, int cap, int weight) {        edges.push_back(Edge(from, to, cap, weight));        edges.push_back(Edge(to, from, 0, -weight));        G[from].push_back(edges.size()-2);        G[to].push_back(edges.size()-1);    }    void init() {        edges.clear();        for(int i = 0; i < maxn; i ++) G[i].clear();    }    bool bellmanFord(int s, int t, int& flow, long long& cost) {        memset(inq, 0, sizeof(inq));        for(int i = 0; i < maxn; i ++) d[i] = INF;        inq[s] = 1;        d[s] = 0;        a[s] = INF;        queue<int> q;        q.push(s);        while(!q.empty()) {            int x = q.front(); q.pop();            inq[x] = 0;            for(int i = 0; i < G[x].size(); i ++) {                Edge& e = edges[G[x][i]];                int y = e.to;                if(e.cap > e.flow && d[y] > d[x] + e.weight) {                    d[y] = d[x] + e.weight;                    p[y] = G[x][i];                    a[y] = min(a[x], e.cap-e.flow);                    if(!inq[y]) { q.push(y); inq[y] = 1; }                }            }        }        if(d[t] == INF) {            return false;        }        for(int u = t; u != s; u = edges[p[u]].from) {            edges[p[u]].flow += a[t];            edges[p[u]^1].flow -= a[t];        }        cost += (long long)d[t]*(long long)a[t];        flow += a[t];        return true;    }    int mincostMaxflow(int s, int t, long long& cost) {        int flow = 0;        cost = 0;        while(flow < 2 && bellmanFord(s, t, flow, cost));        return flow;    }};int main() {    while(scanf("%d%d", &v, &e) != EOF) {        MCMF mcmf;        mcmf.init();        for(int i = 2; i < v; i ++)            mcmf.addEdge(i, i+v, 1, 0);        int a, b, c;        for(int i = 0; i < e; i ++) {            scanf("%d%d%d", &a, &b, &c);            int from = (a==1)?a:a+v, to = b;            mcmf.addEdge(from, to, 1, c);        }        long long cost = 0;        mcmf.mincostMaxflow(1, v, cost);        cout<<cost<<endl;    }    return 0;}

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