题意：

给定一张有向图，求最短路的条数，如果次短路长度 = 最短路 + 1，则输出最短路和次短路条数的和。

思路：

1. 一开始想到 POJ 2449 求 k 短路的方法求解，case 都过的差不多了，无奈 TLE 了，于是找到题解，求最短路的过程中顺带求出次短路;

2. 分别设置二维数组 d[0][i] 表示 s 到 i 的最短路，d[1][i] 表示 s 到 i 的次短路，cnt[][] , vis[][] 数组都是类似的;

3. 依然有地方需要注意的是，当更新某一点最短路/次短路的时候，计数的数组要赋值为其父节点引向其的技术值，具体见代码;

// 方法 1 （dijkstra 47ms)#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;const int MAXN = 1010;
const int INFS = 0x3FFFFFFF;struct edge {int to, cost;edge(int _to, int _cost) : to(_to), cost(_cost) {}
};vector<edge> G[MAXN];
int d[2][MAXN], cnt[2][MAXN];
bool vis[2][MAXN];struct ST {int u, dd, r;ST(int _u, int _dd, int _r) : u(_u), dd(_dd), r(_r) {}bool operator < (const ST& o) const { return dd > o.dd; }
};int dijkstra(int s, int t, int n) {for (int i = 1; i <= n; i++) {d[0][i] = d[1][i] = INFS;cnt[0][i] = cnt[1][i] = 0;vis[0][i] = vis[1][i] = false;}priority_queue<ST> Q;Q.push(ST(s, 0, 0));d[0][s] = 0, cnt[0][s] = 1;while (!Q.empty()) {ST o = Q.top(); Q.pop();int u = o.u, r = o.r;if (vis[r][u]) continue;else vis[r][u] = true;for (int i = 0; i < G[u].size(); i++) {edge& e = G[u][i];int dis = o.dd + e.cost;if (dis < d[0][e.to]) {if (d[0][e.to] != INFS) {cnt[1][e.to] = cnt[0][e.to];d[1][e.to] = d[0][e.to];Q.push(ST(e.to, d[1][e.to], 1));}d[0][e.to] = dis;cnt[0][e.to] = cnt[r][u];Q.push(ST(e.to, dis, 0));}else if (dis == d[0][e.to]) {cnt[0][e.to] += cnt[r][u];} else if (dis < d[1][e.to]) {d[1][e.to] = dis;cnt[1][e.to] = cnt[r][u];Q.push(ST(e.to, dis, 1));} else if (dis == d[1][e.to]) {cnt[1][e.to] += cnt[r][u];}}}int ans = cnt[0][t];if (d[0][t] == d[1][t] - 1)ans += cnt[1][t];return ans;
}int main() {int cases;scanf("%d", &cases);while (cases--) {int n, m;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)G[i].clear();for (int i = 0; i < m; i++) {int u, v, cost;scanf("%d%d%d", &u, &v, &cost);G[u].push_back(edge(v, cost));}int s, t;scanf("%d%d", &s, &t);printf("%d\n", dijkstra(s, t, n));}return 0;
}

// 方法 2：（K 短路解法 TLE）#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;const int MAXN = 1010;
const int INFS = 0x3FFFFFFF;struct edge {int to, cost;edge(int _to, int _cost) : to(_to), cost(_cost) {}
};struct ST {int u, cost, f;ST(int _u, int _cost, int _f): u(_u), cost(_cost), f(_f) {}bool operator < (const ST& other) const { return f > other.f; }
};vector<edge> G[MAXN], P[MAXN];
int d[MAXN];
bool vis[MAXN];void SPFA(int s, int n) {for (int i = 1; i <= n; i++)d[i] = INFS, vis[i] = false;queue<int> Q;Q.push(s); d[s] = 0, vis[s] = true;while (!Q.empty()) {int x = Q.front(); Q.pop();vis[x] = false;for (int i = 0; i < P[x].size(); i++) {edge& e = P[x][i];if (d[e.to] > d[x] + e.cost) {d[e.to] = d[x] + e.cost;if (!vis[e.to]) {vis[e.to] = true; Q.push(e.to);}}}}
}int solve(int s, int t, int n) {priority_queue<ST> Q;Q.push(ST(s, 0, d[t]));int count = 0;bool flag = false;while (!Q.empty()) {ST o = Q.top(); Q.pop();if (o.u == t) {if (o.cost == d[s])count += 1;else if (o.cost == d[s] + 1)count += 1;elsereturn count;}for (int i = 0; i < G[o.u].size(); i++) {edge& e = G[o.u][i];Q.push(ST(e.to, e.cost + o.cost, e.cost + o.cost + d[e.to]));}}return count;
}int main() {int cases;scanf("%d", &cases);while (cases--) {int n, m, s, t;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)G[i].clear(), P[i].clear();for (int i = 0; i < m; i++) {int u, v, cost;scanf("%d%d%d", &u, &v, &cost);G[u].push_back(edge(v, cost));P[v].push_back(edge(u, cost));}scanf("%d%d", &s, &t);SPFA(t, n);printf("%d\n", solve(s, t, n));}return 0;
}