HDU - 3416 Marriage Match IV （最大流）

题目大意：有个人要从A城市去B城市，每条路只允许走一次，问能走几次最短路

解题思路：这题的话，难点就是怎么知道是不是最短路了

首先，先求出到B最短路，这也顺便求出了所有点到B的最短距离

接着求出到A的最短路

这样就能得到两个数组了,假设d1[u]代表u节点到A城市的最短路d2[v]代表v节点到城市B的最短距离

如果满足d1[u] + dis[u][v] + d2[v] == d1[v]的话，那么u，v这条路就属于最短路中的一条边了，那样就可以构边了

得将城市拆成两个点，容量为1

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge {
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}
};

struct ISAP {
    int p[N], num[N], cur[N], d[N];
    int t, s, n, m;
    bool vis[N];

    vector<int> G[N];
    vector<Edge> edges;

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

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

    bool BFS() {
        memset(vis, 0, sizeof(vis));

        queue<int> Q;
        d[t] = 0;
        vis[t] = 1;
        Q.push(t);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i] ^ 1];
                if (!vis[e.from] && e.cap > e.flow) {
                    vis[e.from] = true;
                    d[e.from] = d[u] + 1;
                    Q.push(e.from);
                }
            }
        }
        return vis[s];
    }

    int Augment() {
        int u = t, flow = INF;
        while (u != s) {
            Edge &e = edges[p[u]];
            flow = min(flow, e.cap - e.flow);
            u = edges[p[u]].from;
        }

        u = t;
        while (u != s) {
            edges[p[u]].flow += flow;
            edges[p[u] ^ 1].flow -= flow;
            u = edges[p[u]].from;
        }
        return flow;
    }

    int Maxflow(int s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        BFS();
        if (d[s] >= n)
            return 0;

        memset(num, 0, sizeof(num));
        memset(cur, 0, sizeof(cur));
        for (int i = 0; i < n; i++)
            if (d[i] < INF)
                num[d[i]]++;
        int u = s;

        while (d[s] < n) {
            if (u == t) {
                flow += Augment();
                u = s;
            }
            bool ok = false;
            for (int i = cur[u]; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[u] == d[e.to] + 1) {
                    ok = true;
                    p[e.to] = G[u][i];
                    cur[u] = i;
                    u = e.to;
                    break;
                }
            }

            if (!ok) {
                int Min = n - 1;
                for (int i = 0; i < G[u].size(); i++) {
                    Edge &e = edges[G[u][i]];
                    if (e.cap > e.flow)
                        Min = min(Min, d[e.to]);
                }
                if (--num[d[u]] == 0)
                    break;
                num[d[u] = Min + 1]++;
                cur[u] = 0;
                if (u != s)
                    u = edges[p[u]].from;
            }
        }
        return flow;
    }
};

ISAP isap;
#define M 100010
int d1[N], d2[N], n, m, s, t, cnt;
int head[N], Next[M], dis[M], to[M];
bool vis[N];

void SPFA(int s, int *d) {
    for (int i = 1; i <= n; i++) {
        d[i] = INF;
        vis[i] = 0;
    }

    d[s] = 0;
    queue<int> q;
    q.push(s);

    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;

        for (int i = head[u]; i != -1; i = Next[i]) {
            int v = to[i];
            if (d[v] > d[u] + dis[i]) {
                d[v] = d[u] + dis[i];
                if (!vis[v]) {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
}

void add_edges(int u, int v, int d) {
    to[cnt] = v;
    dis[cnt] = d;
    Next[cnt] = head[u];
    head[u] = cnt++;
}

struct Node {
    int x, y, z;
}node[M];

void solve() {
    scanf("%d%d", &n, &m);

    cnt = 0;
    memset(head, -1, sizeof(head));
    int x, y, z;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &x, &y, &z);
        node[i].x = x;
        node[i].y = y;
        node[i].z = z;
        if (x == y)
            continue;
        add_edges(y, x, z);
    }

    scanf("%d%d", &s, &t);
    SPFA(t, d2);
    cnt = 0;
    memset(head, -1, sizeof(head));
    for (int i = 0; i < m; i++) {
        if (node[i].x == node[i].y)
            continue;
        add_edges(node[i].x, node[i].y, node[i].z);
    }
    SPFA(s, d1);

    isap.init(n);
    for (int i = 1; i <= n; i++)
        for (int j = head[i]; j != -1; j = Next[j]) {
            int v = to[j];
            if (d1[i] + dis[j] + d2[v] == d1[t]) {
                isap.AddEdge(i, v, 1);
            }
        }
    printf("%d\n", isap.Maxflow(s, t));
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        solve();
    }
    return 0;
}

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