HDU - 3416 Marriage Match IV 求最短路条数

题意：n个点m条边， 求a到b最短路条数。

题解：求出a点到所有点的距离， b点到所有点距离， a到b最短路。建一个图， 枚举所有的边，如果此条边在最短路上（dist[a] + dist[b] + wab == mindistab）则加入图中， 容量为1， 超级汇点s连a ，t连b，用dinic 跑一边最大流即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>

using namespace std;

const int MaxN = 2e4;
const int MaxM = 2222222;

struct node
{
    int v, w, next;
}e[MaxM];

struct Node
{
    int v, rev;
    int cap;
    Node(int v, int cap, int rev):v(v), cap(cap), rev(rev){};
};

int n, m,  cnt;
int g[MaxN + 1], a[MaxN + 1];
int dist[MaxN + 1], d1[MaxN + 1], d2[MaxN + 1];
int u[MaxM + 1], v[MaxM + 1], w[MaxM + 1];
vector<Node> f[MaxN + 1];
int cur[MaxN + 1];

void addedge(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = g[u];
    g[u] = cnt++;
}

void add(int u, int v, int cap)
{
    f[u].push_back(Node(v, cap, f[v].size()));
    f[v].push_back(Node(u, 0, f[u].size() - 1));
}

void spfa(int s)
{
    memset(dist, 0x3f, sizeof(dist));
    dist[s] = 0;
    bool vis[MaxN + 1] = {0};
    vis[s] = true;
    queue<int> q;
    q.push(s);

    while (!q.empty())
    {
        int t = q.front(); q.pop();
        vis[t] = false;
        for (int i = g[t]; i != -1; i = e[i].next)
        {
            int v = e[i].v;
            if (dist[v] > dist[t] + e[i].w)
            {
                dist[v] = dist[t] + e[i].w;
                if (!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
}

bool bfs(int s, int t)
{
    memset(dist, -1, sizeof(dist));
    queue<int> q;
    dist[s] = 0;
    q.push(s);

    while (!q.empty())
    {
        int x = q.front(); q.pop();
        if (x == t)
            return true;
        for (int i = 0; i < f[x].size(); i++)
        {
            Node &e = f[x][i];
            if (e.cap > 0 && dist[e.v] < 0)
            {
                dist[e.v] = dist[x] + 1;
                q.push(e.v);
            }
        }
    }
    return false;
}

int dfs(int u, int v, int flow)
{
    if (u == v)
        return flow;

    for (int &i = cur[u]; i < f[u].size(); i++)
    {
        Node &e = f[u][i];
        int d;
        if (e.cap > 0 && dist[u] + 1 == dist[e.v] && (d = dfs(e.v, v, min(flow, e.cap))) > 0)
        {
            e.cap -= d;
            f[e.v][e.rev].cap += d;
            return d;
        }
    }
    return 0;
}

int maxflow(int s, int t)
{
    int flow = 0, f;
    while (bfs(s, t))
    {
        memset(cur, 0, sizeof(cur));
        while ((f = dfs(s, t, 0x3f3f3f3f)) > 0)
            flow += f;
    }
    return flow;
}

int main()
{
    int k = 0;
    scanf("%d", &k);
    while (k--)
    {
        scanf("%d %d", &n, &m);
        memset(g, -1, sizeof(g));
        cnt = 0;
        for (int i = 1; i <= m; i++)
        {
            scanf("%d %d %d", &u[i], &v[i], &w[i]);
            if (u[i] == v[i]) continue;
            addedge(u[i], v[i], w[i]);
        }
        int s, t;
        scanf("%d %d", &s, &t);
        spfa(s);
        int ans = dist[t];
        for (int i = 1; i <= n; i++)
            d1[i] = dist[i];
        memset(g, -1, sizeof(g));
        cnt = 0;
        for (int i = 1; i <= m; i++)
            if (v[i] != u[i])
                addedge(v[i], u[i], w[i]);
        spfa(t);
        for (int i = 1; i <= n; i++)
            d2[i] = dist[i];

        for (int i = 0; i <= n + 1; i++)
            f[i].clear();
        for (int i = 1; i <= m; i++)
            if (d1[u[i]] + w[i] + d2[v[i]] == ans)
                add(u[i], v[i], 1);
        int S = 0, T = n + 1;
        add(S, s, 0x3f3f3f3f);
        add(t, T, 0x3f3f3f3f);
        printf("%d\n", maxflow(S, T));
    }

    return 0;
}