【luogu P1993 小K的农场】 题解

题目链接:https://www.luogu.org/problemnew/show/P1993
1.差分约束:
对于a - b <= c 有一条 b——>a 权值为c
对于a - b >= c 有一条 a——>b 权值为-c
对于一个差分约束系统,若存在负环则无解
2.dfs版SPFA判负环

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ri register
using namespace std;
const int maxn = 100000 + 10;
const int inf = 0x7fffffff;
int n, m, dis[maxn], vis[maxn];
queue<int> q;
struct edge{
    int from, to, next, len;
}e[maxn<<2];
int head[maxn], cnt;
void add(int u, int v, int w)
{
    e[++cnt].from = u;
    e[cnt].len = w;
    e[cnt].next = head[u];
    e[cnt].to = v;
    head[u] = cnt;
}
bool SPFA(int now)
{
    vis[now] = 1;
    for(int i = head[now]; i != -1; i = e[i].next)
    {
        if(dis[e[i].to] > dis[now] + e[i].len)
        {
            dis[e[i].to] = dis[now] + e[i].len;
            if(!vis[e[i].to])
            {
                if(!SPFA(e[i].to)) return 0;
            }
            else return 0;
        }
    }
    vis[now] = 0;
    return 1;
}
int main()
{
    memset(head, -1, sizeof(head));
    scanf("%d%d",&n,&m);
    for(ri int i = 1; i <= n; i++) 
    {
        add(0, i, 0);dis[i] = inf;
    }
    for(ri int i = 1; i <= m; i++)
    {
        int u, v, w, opt;
        scanf("%d",&opt);
        if(opt == 1)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u, v, -w);
        }
        if(opt == 2)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(v, u, w);
        }
        if(opt == 3)
        {
            scanf("%d%d",&u,&v);
            add(v, u, 0);
            add(u, v, 0);
        }
    }
    if(!SPFA(0))
    printf("No\n");
    else
    printf("Yes\n");
    return 0;
}

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转载自www.cnblogs.com/MisakaAzusa/p/9419704.html
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