Description
问题描述:链接
思路
裸题,判断图是否有负环,用 bellman_ford 或者 spfa 。
#include<iostream> #include<algorithm> #include<vector> #include<queue> #include<cstring> using namespace std; #define INF 0x3f3f3f3f int N, M, W; //顶点数 正环数 负权边数 const int MAX_N = 505; struct Edge { int to; int T; Edge(int tt, int TT) : to(tt), T(TT) {} }; vector<Edge> G[MAX_N]; void addEdge (const int& u, const int& v, const int& T) { G[u].push_back(Edge(v, T)); } int dis[MAX_N]; bool inQueue[MAX_N]; int cnt[MAX_N]; bool spfa (const int& s) { memset(dis, INF, sizeof(dis)); memset(inQueue, false, sizeof(inQueue)); memset(cnt, 0, sizeof(cnt)); dis[s] = 0; queue<int> q; q.push(s); inQueue[s] = true; while(!q.empty()) { int u = q.front(); q.pop(); inQueue[u] = false; for (int j = 0; j < G[u].size(); ++j) { int v = G[u][j].to; int cost = G[u][j].T; if (dis[v] > dis[u] + cost ) { dis[v] = dis[u] + cost; if (!inQueue[v]) { q.push(v); inQueue[v] = true; if (++cnt[v] > N) return false; } } } } return true; } int main(void) { int F; cin >> F; while (F--) { cin >> N >> M >> W; for (int i = 1; i <= N; i++) G[i].clear(); for (int i = 1; i <= M; i++) { int S, E, T; cin >> S >> E >> T; addEdge (S, E, T); addEdge (E, S, T); } for (int i = 1; i <= W; i++) { int S, E, T; cin >> S >> E >> T; addEdge (S, E, -1 * T); } if (!spfa(1)) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }