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POJ 3259 最短路 判负环

2013-08-18 18:15 183 查看
SPFA 判负环 某个点访问n次

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef  long long LL;
const double PI = acos(-1.0);

template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}
template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}

const int N = 111;
const int M = 11111;
const LL MOD = 1000000007LL;
const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
const int INF = 0x3f3f3f3f;

int dist[555];
struct node
{
int v, w, next;
}edge[6666];

int head[555], cnt;
int n, m, vis[555];
int inq[555];

inline void addnode(int u, int v, int w)
{
edge[cnt].v = v; edge[cnt].next = head[u];
edge[cnt].w = w; head[u] = cnt++;
}

bool solve()
{
queue < int > q;
fill(dist, dist + n + 1, INF);
memset(vis, 0, sizeof(vis));
memset(inq, 0, sizeof(inq));
dist[1] = 0; vis[1]++; inq[1]++;
int u, i, j, k, v;
q.push(1);
while (!q.empty())
{
u = q.front(); q.pop(); inq[u]--;
for (i = head[u]; ~i; i = edge[i].next)
{
v = edge[i].v;
if (dist[u] + edge[i].w < dist[v])
{
dist[v] = dist[u] + edge[i].w;
if (!inq[v]) {q.push(v); vis[v]++; inq[v]++; if (vis[v] >= n) return true;}
}
}
}
return false;
}

int main()
{
int T;
scanf("%d", &T);
while (T--)
{
int u, v, w, k, i;
memset(head, -1, sizeof(head));
cnt = 0;
scanf("%d%d%d", &n, &m, &k);
for (i = 0; i < m; ++i)
{
scanf("%d%d%d", &u, &v, &w);
addnode(u, v, w);
addnode(v, u, w);
}
for (i = 0; i < k; ++i)
{
scanf("%d%d%d", &u, &v, &w);
addnode(u, v, -w);
}
int flag = 0;
if (solve()) printf("YES\n");
else printf("NO\n");
}
return 0;
}


 
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