【题目链接】
https://www.lydsy.com/JudgeOnline/problem.php?id=1975
【算法】
A*求k短路
【代码】
#include<bits/stdc++.h> using namespace std; #define MAXN 5010 #define MAXM 200010 const double INF = 1e15; int i,tot,n,m,u,v; int head[MAXN],rhead[MAXN]; double dist[MAXN]; double w,val; struct Edge { int to; double w; int nxt; } e[MAXM<<1]; struct info { int s; double d; friend bool operator < (info a,info b) { return a.d + dist[a.s] > b.d + dist[b.s]; } } ; inline void add(int u,int v,double w) { tot++; e[tot] = (Edge){v,w,head[u]}; head[u] = tot; tot++; e[tot] = (Edge){u,w,rhead[v]}; rhead[v] = tot; } inline void dijkstra(int s) { int i,u,v; double w; priority_queue< pair<double,int> > q; static bool vis[MAXN]; memset(vis,false,sizeof(vis)); while (!q.empty()) q.pop(); for (i = 1; i <= n; i++) dist[i] = INF; dist[s] = 0; q.push(make_pair(0,s)); while (!q.empty()) { u = q.top().second; q.pop(); if (vis[u]) continue; vis[u] = true; for (i = rhead[u]; i; i = e[i].nxt) { v = e[i].to; w = e[i].w; if (dist[u] + w < dist[v]) { dist[v] = dist[u] + w; q.push(make_pair(-dist[v],v)); } } } } inline int Astar(int s,int t) { int i,cnt = 0,v; double w,sum = 0; priority_queue< info > q; info cur; while (!q.empty()) q.pop(); q.push((info){s,0}); while (!q.empty()) { cur = q.top(); q.pop(); if (cur.s == t) { if (sum + cur.d <= val) { sum += cur.d; cnt++; } else return cnt; } for (i = head[cur.s]; i; i = e[i].nxt) { v = e[i].to; w = e[i].w; q.push((info){v,cur.d+w}); } } return 0; } int main() { scanf("%d%d%lf",&n,&m,&val); for (i = 1; i <= m; i++) { scanf("%d%d%lf",&u,&v,&w); add(u,v,w); } dijkstra(n); printf("%d\n",Astar(1,n)); return 0; }