本文主要是介绍poj 3255 次短路（第k短路） A* + spfa 或 dijkstra，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题意：
给一张无向图，求从1到n的次短路。
解析：
A* + spfa 或者 dijkstra。
详解见上一题：http://blog.csdn.net/u013508213/article/details/46400189
本题，spfa中，stack超时，queue的效率最高，priority_queue次之。
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long
#define lson lo, mi, rt << 1
#define rson mi + 1, hi, rt << 1 | 1using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 5000 + 1;
const int maxm = 100000 + 1;int n, m;
int st, ed;
int edgeNum;
int head[maxn];struct Edge
{int to, cost, nxt;
} e[maxm << 1];void initEdge()
{memset(head, -1, sizeof(head));edgeNum = 0;
}void addEdge(int fr, int to, int cost)
{e[edgeNum].to = to;e[edgeNum].cost = cost;e[edgeNum].nxt = head[fr];head[fr] = edgeNum++;
}int dis[maxn];
bool vis[maxn];
///stack TLE queue AC priority_queue AC
void spfa()
{for (int i = 1; i <= n; i++){dis[i] = inf;vis[i] = false;}priority_queue<int> q;q.push(ed);vis[ed] = true;dis[ed] = 0;while (!q.empty()){int cur = q.top();q.pop();for (int i = head[cur]; i != -1; i = e[i].nxt){int x = e[i].to;if (dis[cur] + e[i].cost < dis[x]){dis[x] = dis[cur] + e[i].cost;if (!vis[x]){vis[x] = true;q.push(x);}}}vis[cur] = false;}
}//void dij()
//{
// int i,j,k;
// bool vis[maxn];
// memset (vis,false,sizeof(vis));
// memset (dis,0x3f,sizeof(dis));
// dis[ed] = 0;
// for (i = 1; i <= n; i ++)
// {
// int min = inf;
// for (j = 1; j <= n; j ++)
// if (!vis[j]&&dis[j] < min)
// {
// min = dis[j];
// k = j;
// }
// vis[k] = true;
// for (int u = head[k]; u != -1; u = e[u].nxt)
// {
// int v = e[u].to;
// if (!vis[v]&&dis[v] > dis[k] + e[u].cost)
// dis[v] = dis[k] + e[u].cost;
// }
// }
//}struct Node
{int pos;int h, g;bool operator < (const Node a) const{return a.h + a.g < h + g;}
};int Astar(int k)
{int cnt = 0;if (st == ed)k++;if (dis[st] == inf)return -1;priority_queue<Node> q;Node now, nxt;now.pos = st, now.g = 0, now.h = dis[st];q.push(now);while (!q.empty()){nxt = q.top();q.pop();if (nxt.pos == ed){cnt++;if (cnt == k)return nxt.g;}for (int i = head[nxt.pos]; i != -1; i = e[i].nxt){now.pos = e[i].to;now.g = nxt.g + e[i].cost;now.h = dis[e[i].to];q.push(now);}}return 0;
}int main()
{
#ifdef LOCALfreopen("in.txt", "r", stdin);
#endif // LOCALwhile (~scanf("%d%d", &n, &m)){initEdge();while (m--){int fr, to, cost;scanf("%d%d%d", &fr, &to, &cost);addEdge(fr, to, cost);addEdge(to, fr, cost);}st = 1;ed = n;spfa();
// dij();printf("%d\n", Astar(2));}return 0;
}
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