poj 3255 次短路(第k短路) A* + spfa 或 dijkstra
阿新 • • 發佈:2019-01-06
題意:
給一張無向圖,求從1到n的次短路。
解析:
A* + spfa 或者 dijkstra。
本題,spfa中,stack超時,queue的效率最高,priority_queue次之。
程式碼:
#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 | 1 using 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 LOCAL freopen("in.txt", "r", stdin); #endif // LOCAL while (~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; }