67 lines
2.0 KiB
C++
67 lines
2.0 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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//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
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const int N = 110, M = N * N * 100;
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//邻接表
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int d[N]; //最短距离数组
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bool st[N]; //是不是进过队列
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int cnt[N];
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int n, m, T;
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bool spfa(int start) { //求最长路，所以判断正环
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queue<int> q; //有时候换成栈判断环很快就能
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//差分约束从超级源点出发
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d[start] = 0;
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q.push(start);
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st[start] = true;
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while (q.size()) {
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int t = q.front();
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q.pop();
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st[t] = false;
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for (int i = h[t]; ~i; i = ne[i]) {
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int j = e[i];
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if (d[j] > d[t] + w[i]) { //求最短路
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d[j] = d[t] + w[i];
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cnt[j] = cnt[t] + 1;
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//注意多加了超级源点到各各节点的边
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if (cnt[j] >= n + 1) return false;
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if (!st[j]) {
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q.push(j);
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st[j] = true;
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}
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}
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}
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}
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return true;
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}
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int main() {
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cin >> T;
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while (T--) {
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cin >> n >> m;
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memset(h, -1, sizeof h);
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memset(d, 0x3f, sizeof(d));
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memset(cnt, 0, sizeof(cnt));
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memset(st, 0, sizeof(st));
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while (m--) {
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int s, t, v;
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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
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add(t, s - 1, -v), add(s - 1, t, v);
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}
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//建立超级源点
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for (int i = 1; i <= n; i++) add(n + 1, i, 0);
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//整条路径中存在负环，意味着不等式组存在矛盾
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if (!spfa(n + 1))
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cout << "false" << endl;
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else
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cout << "true" << endl;
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}
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return 0;
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} |