HDU ~ 3416 ~ Marriage Match IV （最短路Dijkstra + 网络流Dinic）

题意：T组测试数据，给你N个点M条单向边，再给你起点A，终点B， 问你A~B的最短路最多有多少条(每条边只能走一次)

思路：先求出在最短路上的所有的边，然后以这些边建图，容量均为1，跑A~B的最大流就是答案。

正向建边跑一边最短路得到一个以A为源点最短路（dis1），反向建边再跑一次得到一个以B为起点的最短路（dis2），最短路径长度为dis1[B]（dis1[B]==dis2[A]）。如果对于边权为w的边u~v，如果dis1[u] + w + dis2[v] == dis1[B]，那么这条边一定在最短路上。遍历所有的边，以所有最短路上的边建图，跑一波A~B的最大流就OK。

代码较长，变量名较多，记得初始化，别写出BUG就好。我第一次自己手敲这么长的代码，结果调试了几分钟就AC了，虽然题比较简单，而且两个算法都是模板，但是还是很开心的，哈哈哈！

Dijkstra + Dinic：

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 5;
const int MAXM = 1e5 + 5;
const int INF = 0x3f3f3f3f;

//Dijkstra
struct Edge1
{
    int from, to, dist;
    Edge1(int u, int v, int w): from(u), to(v), dist(w) {}
};

struct Dijkstra
{
    int n, m;
    vector<Edge1> edges;
    vector<int> G[MAXN];
    int d[MAXN];
    int vis[MAXN];
    int pre[MAXN];

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

    void add_edge(int from, int to, int dist)
    {
        edges.push_back(Edge1(from, to, dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }

    struct HeapNode
    {
        int from, dist;
        bool operator < (const HeapNode& rhs) const
        {
            return rhs.dist < dist;
        }
        HeapNode(int u, int w): from(u), dist(w) {}
    };
    void dijkstra(int s)
    {
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        priority_queue<HeapNode> Q;
        Q.push(HeapNode(s, 0));
        while (!Q.empty())
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.from;
            if (vis[u]) continue;
            vis[u] = true;
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge1& e = edges[G[u][i]];
                if (!vis[e.to] && d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    pre[e.to] = G[u][i];
                    Q.push(HeapNode(e.to, d[e.to]));
                }
            }
        }
    }
};

//Dinic
struct Edge2
{
    int from, to, cap, flow;
    Edge2(int u, int v, int c, int f): from(u), to(v), cap(c), flow(f) {}
};

struct Dinic
{
    int n, m, s, t;
    vector<Edge2> edges;
    vector<int> G[MAXN];
    int d[MAXN];
    int vis[MAXN];
    int cur[MAXN];

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

    void add_edge(int from, int to, int cap)
    {
        edges.push_back(Edge2(from, to, cap, 0));
        edges.push_back(Edge2(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));
        d[s] = 0; vis[s] = true;
        queue<int> Q;
        Q.push(s);
        while (!Q.empty())
        {
            int u = Q.front(); Q.pop();
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge2& e = edges[G[u][i]];
                if (!vis[e.to] && e.cap > e.flow)
                {
                    d[e.to] = d[u] + 1;
                    vis[e.to] = true;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a)
    {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int& i = cur[x]; i < G[x].size(); i++)
        {
            Edge2& e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = DFS(e.to,  min(a, e.cap - e.flow))) > 0)
            {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    int max_flow(int s, int t)
    {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS())
        {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }
};

struct Edge
{
    int from, to, dist;
}edge[MAXM];

int main()
{
    int T, n, m, A, B;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        Dijkstra solve1, solve2;//正反向最短路
        solve1.init(n); solve2.init(n);
        for (int i = 0; i < m; i++)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            edge[i].from = u; edge[i].to = v; edge[i].dist= w;
            solve1.add_edge(u, v, w);
            solve2.add_edge(v, u, w);
        }
        scanf("%d%d", &A, &B);
        solve1.dijkstra(A);
        solve2.dijkstra(B);
        int ans = solve1.d[B];//最短路径长度

        Dinic solve3;
        solve3.init(n);
        for (int i = 0; i < m; i++)
        {
            Edge e = edge[i];
            if (solve1.d[e.from] + e.dist + solve2.d[e.to] == ans)
            {
                solve3.add_edge(e.from, e.to, 1);
            }
        }
        ans = solve3.max_flow(A, B);//A~B的最大流
        printf("%d\n", ans);
    }
    return 0;
}
/*
3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2
*/