洛谷P3385 【模板】负环 题解

我的第一道洛谷题：^_^

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 思路:

1.初始化（不然会有悲剧发生，我就经历过）

memset(head, 0, sizeof(head));
memset(exsit, 0, sizeof(exsit));
memset(d, 0x3f, sizeof(d));
memset(count, 0, sizeof(count));

2.建图（非常简单的一步~）

inline void addEdge(int x, int y, int z)
{
    G[++num].to=y;
    G[num].data=z;
    G[num].next=head[x];
    head[x]=num;
}

3.spfa（简单模板）

inline bool spfa(int x)
{
    d[x] = 0;
    exsit[x] = 1;
    q.push(x);
    count[x] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        exsit[u] = 0;
        for(int i=head[u]; i; i = G[i].next)
        {
            int v = G[i].to;
            int w = G[i].data;
            if(d[v] > d[u] + w)
            {
                d[v] = d[u] + w;
                if(!exsit[v])
                {
                    count[v]++;
                    if(count[v] > n-1)
                        return false;
                    exsit[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return true;
}

有没有感觉异常简单呢？只是一道基础模板题而已。只是输出上有点小坑，一定要看清题啊！！！！！！！！！

---

贴个完整代码吧：

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <cstdio>
#define F(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int MAXN = 6000;
const int MAXM = 60000;

int n, m, num, T;
int head[MAXN], exsit[MAXN], d[MAXN], count[MAXN];
queue <int> q;

struct Edge
{
    int to, data, next;

}G[MAXM];

inline void addEdge(int x, int y, int z)
{
    G[++num].to=y;
    G[num].data=z;
    G[num].next=head[x];
    head[x]=num;
}

inline bool spfa(int x)
{
    d[x] = 0;
    exsit[x] = 1;
    q.push(x);
    count[x] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        exsit[u] = 0;
        for(int i=head[u]; i; i = G[i].next)
        {
            int v = G[i].to;
            int w = G[i].data;
            if(d[v] > d[u] + w)
            {
                d[v] = d[u] + w;
                if(!exsit[v])
                {
                    count[v]++;
                    if(count[v] > n-1)
                        return false;
                    exsit[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return true;
}

int main()
{
    cin >> T;
    while(T--)
    {
        memset(head, 0, sizeof(head));
        memset(exsit, 0, sizeof(exsit));
        memset(d, 0x3f, sizeof(d));
        memset(count, 0, sizeof(count));
        while(!q.empty())
            q.pop();
        n = read();
        m = read();
        int a, b, c;
        F(i,1,m)
        {
            a = read();
            b = read();
            c = read();
            if(c < 0)
            {
                addEdge(a, b, c);
            }
            else
            {
                addEdge(a, b, c);
                addEdge(b, a, c);
            }
        }
        if(!spfa(1))
            cout << "YE5" << endl;
        else cout << "N0" << endl;
    }
    return 0;
}

最后， 别忘了开o2，最好用Bellman_Ford哦~(有没有感觉被骗了,逃