HDU 6181:Two Paths(A* + SPFA)
题意
给出n个点m条边的无向图,求次短路。
思路
和 POJ 2449 类似,只不过大小要开成long long。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100011;
const int INF = 0x3f;
struct Edge {
int v, nxt; LL w;
} edge[N*2];
struct Node {
int u; LL g, h;
friend bool operator < (const Node &a, const Node &b) {
return a.g + a.h > b.h + b.g;
}
};
int n, m, vis[N], head[N], tot;
LL h[N];
void Add(int u, int v, LL w) {
edge[tot] = (Edge) { v, head[u], w }; head[u] = tot++;
edge[tot] = (Edge) { u, head[v], w }; head[v] = tot++;
}
void Spfa(int s, int t) {
memset(h, INF, sizeof(h));
memset(vis, 0, sizeof(vis));
queue<int> que; que.push(t);
vis[t] = 1; h[t] = 0;
while(!que.empty()) {
int u = que.front(); que.pop();
vis[u] = 0;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v, w = edge[i].w;
if(h[v] > h[u] + w) {
h[v] = h[u] + w;
if(!vis[v]) vis[v] = 1, que.push(v);
}
}
}
}
LL Astar(int s, int t, int k) {
Node now = (Node) { s, 0LL, h[s] };
priority_queue<Node> que; que.push(now);
int cnt = 0;
while(!que.empty()) {
now = que.top(); que.pop();
int u = now.u; LL gg = now.g, hh = now.h;
// printf("%d : %d - %d\n", u, gg, hh);
if(u == t) cnt++;
if(cnt == k) return gg + hh;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v; LL w = edge[i].w;
now = (Node) { v, gg + w, h[v] };
que.push(now);
}
}
return 0;
}
int main() {
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
memset(head, -1, sizeof(head)); tot = 0;
for(int i = 1; i <= m; i++) {
int u, v; LL w;
scanf("%d%d%lld", &u, &v, &w);
Add(u, v, w);
}
Spfa(1, n);
printf("%lld\n", Astar(1, n, 2));
}
return 0;
}