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uva 10806 Dijkstra, Dijkstra. (最小费最大流)

uva 10806 Dijkstra, Dijkstra.

题目大意:你和你的伙伴想要越狱。你的伙伴先去探路,等你的伙伴到火车站后,他会打电话给你(电话是藏在蛋糕里带进来的),然后你就能够跑去火车站了,那里有人接应你。

可是。由于你的伙伴跑去火车站的时候穿的是囚服,所以,他经过的街道都被戒严了,你必须从其它街道跑过去。

假设你能够到达火车站,请输出你和你的伙伴在路上花费的最短时间,假设不能请“Back to jail”。

解题思路:最小费最大流。设置一个超级源点连向监狱(起点1), 容量为2(两个人),设置一个超级汇点,使火车站(终点n)连向他,容量为2(两个人)。其余街道皆为无向(即正向反向都要考虑)。且容量为1(每条街道仅仅能跑一次)。最后求最大流,若最大流为2则输出最小费,否则回监狱准备下次越狱吧。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 105;
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[3 * N];

void init() {
    for (int i = 0; i < 3 * 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() {
    int from, to;
    ll cost;
    for (int i = 0; i < m; i++) {
        scanf("%d %d %lld", &from, &to, &cost); 
        addEdge(from, to, 1, 0, cost);
        addEdge(to, from, 1, 0, cost);
    }   
    addEdge(0, 1, 2, 0, 0);
    addEdge(n, n + 1, 2, 0, 0);
}
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", &n) != EOF) {
        if (n == 0) break;
        scanf("%d", &m);
        s = 0, t = n + 1;
        init();
        input();
        ll cost = 0;
        int ans = MCMF(s, t, cost);
        if (ans == 1) printf("Back to jail\n");
        else printf("%lld\n", cost);
    }   
    return 0;
}
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uva 10806 Dijkstra, Dijkstra. (最小费最大流)