转化题意后就是如果一个点 \(B\) 到终点的最短路小于 \(A\) 到终点的最短路,则 \(A->B\) 可以通过。
从终点求一遍最短路,新图是 \(DAG\),记忆化搜索统计路径即可
#include<bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1010; const int INF = 0x3f3f3f3f; int n, m; int h[maxn], cnt = 0; struct E{ int to, cost, next; }e[maxn * maxn * 2]; void add(int u, int v, int w){ e[++cnt].to = v; e[cnt].cost = w; e[cnt].next = h[u]; h[u] = cnt; } int d[maxn]; void dij(int S){ memset(d, 0x3f, sizeof(d)); priority_queue<pii, vector<pii>, greater<pii> > q; d[S] = 0; q.push(pii(0, S)); while(!q.empty()){ pii p = q.top(); q.pop(); int u = p.second; if(p.first != d[u]) continue; for(int i = h[u] ; i != -1 ; i = e[i].next){ int v = e[i].to; if(d[v] > d[u] + e[i].cost){ d[v] = d[u] + e[i].cost; q.push(pii(d[v], v)); } } } } int dp[maxn]; int DP(int u){ if(dp[u] > 0) return dp[u]; for(int i = h[u] ; i != -1 ; i = e[i].next){ int v = e[i].to; if(d[u] > d[v]) dp[u] += DP(v); } return dp[u]; } ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; } int main(){ while(scanf("%d", &n) == 1 && n){ memset(h, -1, sizeof(h)); cnt = 0; scanf("%d", &m); int u, v, w; for(int i = 1 ; i <= m ; ++i){ scanf("%d%d%d", &u, &v, &w); add(u, v, w), add(v, u, w); } dij(2); memset(dp, 0, sizeof(dp)); dp[2] = 1; DP(1); printf("%d\n", dp[1]); } return 0; }