POJ3259

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);                    }                }            }            q.pop();            vis[u] = false;        }        if (flag)            printf("YES\n");        else            printf("NO\n");    }}