uva 1658 Admiral （最小费最大流）

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uva 1658 Admiral

题目大意：在图中找出两条没有交集的线路，要求这两条线路的费用最小。

解题思路：还是拆点建图的问题。首先每个点都要拆成两个点，例如a点拆成a->a’。起点和终点的两点间的容量为2费用为0，保证了只找出两条线路。其余点的容量为1费用为0，保证每点只走一遍，两条线路无交集。然后根据题目给出的要求继续建图。每组数据读入a, b, c，建立a’到b的边容量为1，费用为c。图建完之后，用bellman-ford来实现MCMF。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <queue>using namespace std;typedef long long ll;const int N = 2005;const int INF = 0x3f3f3f3f;int n, m, s, t;int a[N], pre[N], d[N], inq[N]; struct Edge{    int from, to, cap, flow;    ll cos;};vector<Edge> edges;vector<int> G[N];void init() {    for (int i = 0; i < 2 * n; i++) G[i].clear();    edges.clear();}void addEdge(int from, int to, int cap, int flow, ll cos) {    edges.push_back((Edge){from, to, cap, 0, cos});    edges.push_back((Edge){to, from, 0, 0, -cos});    int m = edges.size();    G[from].push_back(m - 2);    G[to].push_back(m - 1);}void input() {    addEdge(1, n + 1, 2, 0, 0);    for (int i = 2; i <= n - 1; i++) {        addEdge(i, i + n, 1, 0, 0);     }    addEdge(n, 2 * n, 2, 0, 0);    int u, v;    ll c;    for (int i = 0; i < m; i++) {        scanf("%d %d %lld", &u, &v, &c);            addEdge(u + n, v, 1, 0, c);    }}int BF(int s, int t, int& flow, ll& cost) {    queue<int> Q;    memset(inq, 0, sizeof(inq));    memset(a, 0, sizeof(a));    memset(pre, 0, sizeof(pre));    for (int i = 0; i <= 2 * n + 1; i++) d[i] = INF;    d[s] = 0;    a[s] = INF;    inq[s] = 1;    int flag = 1;    pre[s] = 0;    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.cos) {                d[e.to] = d[u] + e.cos;                a[e.to] = min(a[u], e.cap - e.flow);                pre[e.to] = G[u][i];                if (!inq[e.to]) {                    inq[e.to] = 1;                    Q.push(e.to);                }            }           }        flag = 0;    }    if (d[t] == INF) return 0;    flow += a[t];    cost += (ll)d[t] * (ll)a[t];    for (int u = t; u != s; u = edges[pre[u]].from) {        edges[pre[u]].flow += a[t];        edges[pre[u]^1].flow -= a[t];    }    return 1;}int MCMF(int s, int t, ll& cost) {    int flow = 0;    cost = 0;           while (BF(s, t, flow, cost));    return flow;}int main() {    while (scanf("%d %d", &n, &m) == 2) {        s = 1, t = 2 * n;           ll cost;        init();        input();        MCMF(s, t, cost);        printf("%lld\n", cost);    }    return 0;}

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