POJ

来源：互联网发布：蛋鸡存栏量数据编辑：程序博客网时间：2024/06/08 08:18

题意：一个农民发现了一些虫洞，虫洞能够在把他从u带到v的同时，将当前花费时间减少T，问他能否从一个点出发然后利用虫洞和农场之间的道路回到原点并且时间比出发时更小？

思路：虫洞可以看做负边，直接判断是否存在负环即可。但是题目并没有说整个图是连通的，可能是几个联通块？但是当做一个连通图处理也AC了。spfa or bellman-ford or floyd都能做

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 500 + 5;bool inq[maxn];int cnt[maxn], d[maxn];int n, m, w;vector<int>G[maxn];struct Edge{int from, to, dist;Edge(int u, int v, int d):from(u), to(v), dist(d){}};vector<Edge>edges;void init(int n) {edges.clear();for(int i = 0; i <= n; ++i) {d[i] = inf;G[i].clear();}memset(inq, false, sizeof(inq));memset(cnt, 0, sizeof(cnt));}void add_edge(int u, int v, int dis) {edges.push_back(Edge(u, v, dis)); int m = edges.size();G[u].push_back(m-1);}bool spfa(int s) {queue<int>q;d[s] = 0;q.push(s);inq[s] = true;while(!q.empty()) {int u = q.front(); q.pop();inq[u] = false;for(int i = 0; i < G[u].size(); ++i) {Edge &e = edges[G[u][i]];if(d[u] < inf && d[e.to] > d[u] + e.dist) {d[e.to] = d[u] + e.dist;if(!inq[e.to]) {q.push(e.to);inq[e.to] = true;if(++cnt[e.to] > n) return true; //存在负环 }}}}return false;}int main() {int T;scanf("%d", &T);while(T--) {scanf("%d%d%d", &n, &m, &w);init(n);int u, v, dis;for(int i = 0; i < m; ++i) { //无向边 scanf("%d%d%d", &u, &v, &dis);add_edge(u, v, dis);add_edge(v, u, dis);}for(int i = 0; i < w; ++i) { //虫洞--有向边 scanf("%d%d%d", &u, &v, &dis);add_edge(u, v, -dis);}if(spfa(1)) printf("YES\n");else printf("NO\n"); }return 0;}

如有不当之处欢迎指出！

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