bellman_ford算法判负环-洛谷P3371

总结:这题改了很久,一直wa

问题一:多测要清空数组

问题二:本题其实有点特殊,它要求的是,从1开始出发能到达的负环,也就是这个1实际上必须能到这个负环,如果图不连通,就会出现有负环但1去不了,等于没有负环的情况,这种特殊情况bellman_ford算法是压根没法解决的 一个玄学方法就是:判断1是否有出度,但实际上这个玄学方法仅仅能过这题的特例,换一个1有出度到不了的负环就hack住了

#pragma optimize(2)
#include
#include
#include
#define endl '\n'
#define int int64_t
using namespace std;
const int N = 1e5 + 10;
struct edge { int v, w; };
vectore[N];
int d[N],m,n;
bool bellman_ford(int s) {
    memset(d, 0x3f3f3f3f, sizeof d);
    d[s] = 0;
    bool sign;
    for (int i = 1; i <= n; ++i) {//次数
        sign = false;
         for (int u = 1; u <= n; ++u) {//顶点
             if (d[u] == 0x3f3f3f3f)continue;
             for (auto k : e[u]) {
                 if (d[k.v] > d[u] + k.w) {
                       d[k.v] = d[u] + k.w;
                       sign = true;
                   }
             }
        }
         if (!sign) break;
    }
    return sign;
}
signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t,a,b,c; cin >> t;
    while (t--) {
        for (int i = 1; i <= n; ++i) e[i].erase(e[i].begin(), e[i].end());
        cin >> n >> m;
        for (int i = 1; i <= m; ++i) {
            cin >> a >> b >> c;
            e[a].push_back({ b,c });
            if (c >= 0) e[b].push_back({ a,c });
        }
        if (bellman_ford(1)) {
            if (e[1].size()) cout << "YES" << endl;
            else cout << "NO" << endl;
        }
        else cout << "NO" << endl;
    }
    return 0;
}