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2 years ago
#include <bits/stdc++.h>
using namespace std;
//这里的M取值需要总结一下因为有100组以下的输入所以还需要乘以100才是真正的边数上限
const int N = 110, M = N * N * 100;
//邻接表
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int d[N]; //最短距离数组
bool st[N]; //是不是进过队列
int cnt[N];
int n, m, T;
bool spfa(int start) { //求最长路,所以判断正环
queue<int> q; //有时候换成栈判断环很快就能
//差分约束从超级源点出发
d[start] = 0;
q.push(start);
st[start] = true;
while (q.size()) {
int t = q.front();
q.pop();
st[t] = false;
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i];
if (d[j] > d[t] + w[i]) { //求最短路
d[j] = d[t] + w[i];
cnt[j] = cnt[t] + 1;
//注意多加了超级源点到各各节点的边
if (cnt[j] >= n + 1) return false;
if (!st[j]) {
q.push(j);
st[j] = true;
}
}
}
}
return true;
}
int main() {
cin >> T;
while (T--) {
cin >> n >> m;
memset(h, -1, sizeof h);
memset(d, 0x3f, sizeof(d));
memset(cnt, 0, sizeof(cnt));
memset(st, 0, sizeof(st));
while (m--) {
int s, t, v;
cin >> s >> t >> v; //这里c本身就是前缀的结果不需要我们另外计算前缀和
add(t, s - 1, -v), add(s - 1, t, v);
}
//建立超级源点
for (int i = 1; i <= n; i++) add(n + 1, i, 0);
//整条路径中存在负环,意味着不等式组存在矛盾
if (!spfa(n + 1))
cout << "false" << endl;
else
cout << "true" << endl;
}
return 0;
}