[USACO2006 DEC] Wormholes

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1715

[算法]

用SPFA判定负环是否存在即可

时间复杂度 ： O(N ^ 2)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 500010
const int inf = 2e9;

struct edge
{
        int to , w , nxt;
} e[MAXN];

int n , m , p , tot;
int head[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline void addedge(int u , int v , int w)
{
        ++tot;
        e[tot] = (edge){v , w , head[u]};
        head[u] = tot;
}
inline bool spfa()
{
        static int dist[MAXN] , cnt[MAXN];
        static bool inq[MAXN];
        queue< int > q;
        for (int i = ; i <= n; i++)
        {
                cnt[i] = ;
                dist[i] = inf;
                inq[i] = false;
        }
        cnt[] = ;
        dist[] = ;
        q.push();
        while (!q.empty())
        {
                int cur = q.front();
                q.pop();
                inq[cur] = false;
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w;
                        if (dist[cur] + w < dist[v])
                        {
                                dist[v] = dist[cur] + w;
                                if (!inq[v])
                                {
                                        inq[v] = true;
                                        if (++cnt[v] > n) return true;
                                        q.push(v);
                                }
                        }
                }
        }
        return false;
}

int main()
{

        int T;
        read(T);
        while (T--)
        {
                read(n); read(m); read(p);
                tot = ;
                for (int i = ; i <= n; i++) head[i] = ;
                for (int i = ; i <= m; i++)
                {
                        int u , v , w;
                        read(u); read(v); read(w);
                        addedge(u , v , w);
                        addedge(v , u , w);
                }
                for (int i = ; i <= p; i++)
                {
                        int u , v , w;
                        read(u); read(v); read(w);
                        addedge(u , v , -w);
                }
                if (spfa()) printf("YES\n");
                else printf("NO\n");
        }

        return ;

}