POJ3259 - Wormholes(连通图判断负环)
2017-06-13 11:10
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题目链接;
http://poj.org/problem?id=3259题目大意:
给出N个图,每个图有两种边,一个是无向的正权边,一种是有向的负权边,保证所给的图为连通图,求是否存在负环。解题过程:
刚开始以为给出的图不连通,然后用Floyd超时,后来问了学长,翻了下POJ的讨论,发现大家都是默认为图连通做的……然后敲了下Bellman和SPFA判断负环就A了。
题目分析:
因为保证图联通,那么可以假设从任意一点出发。Bellman:如果松弛操进行N次依然可以松弛,那么存在负环。
SPFA:如果一个点入队次数大于等于N次,那么处在负环。
AC代码:
Bellman:
#include <cstdio> #include <cstring> using namespace std; typedef long long LL; struct Node { int u, v, w; }edge[2500*10]; LL dist[1123]; int main() { int f; scanf("%d", &f); while (f--) { int n, m, w; scanf("%d %d %d", &n, &m, &w); for (int i = 0; i < m; i++) { int u, v, c; scanf("%d %d %d", &u, &v, &c); edge[i*2] = {u, v, c}; edge[i*2+1] = {v, u, c}; } m *= 2; for (int i = 0; i < w; i++) { int u, v, c; scanf("%d %d %d", &u, &v, &c); edge[i+m] = {u, v, -c}; } bool flag = false; memset(dist, 0x3f, sizeof(dist)); dist[1] = 0; for (int k = 0; k <= n; k++) { for (int j = 0; j < m + w; j++) { int u = edge[j].u; int v = edge[j].v; int w = edge[j].w; if (dist[v] > dist[u] + w) { if (k == n) flag = true; dist[v] = dist[u] + w; } } } if (flag) printf("YES\n"); else printf("NO\n"); } }
SPFA:
#include <cstdio> #include <cstring> #include <queue> using namespace std; typedef long long LL; vector<pair<int, int> > edge[1123]; int dist[1123], cnt[1123]; bool vis[1123]; int main() { int f; scanf("%d", &f); while (f--) { int n, m, w; scanf("%d %d %d", &n, &m, &w); for (int i = 0; i <= n; i++) { edge[i].clear(); } for (int i = 0; i < m; i++) { int u, v, c; scanf("%d %d %d", &u, &v, &c); edge[u].push_back(make_pair(v, c)); edge[v].push_back(make_pair(u, c)); } for (int i = 0; i < w; i++) { int u, v, c; scanf("%d %d %d", &u, &v, &c); edge[u].push_back(make_pair(v, -c)); } bool flag = false; memset(dist, 0x3f, sizeof(dist)); memset(cnt, 0, sizeof(cnt)); queue<int> q; q.push(1); dist[1] = 0; vis[1] = true; while (!q.empty()) { int u = q.front(); for (int i = 0; i < edge[u].size(); i++) { int v = edge[u][i].first; int w = edge[u][i].second; if (dist[v] > dist[u] + w) { dist[v] = dist[u] + w; if (++cnt[v] >= n) { flag = true; break; } if (!vis[v]) { q.push(v); 4000 } } } q.pop(); vis[u] = false; } if (flag) printf("YES\n"); else printf("NO\n"); } }
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