HDU

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;}