【luogu P2984 [USACO10FEB]给巧克力Chocolate Giving】 题解

题目链接：https://www.luogu.org/problemnew/show/P2984

练习SPFA，把FJ当做起点，求出到所有牛的最短路，再把两个牛的相加。

#include <cstdio>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1000000;
const int inf = 0x7fffffff;
int n, m, b, dis[maxn];
bool vis[maxn];
struct edge{
    int len, to, next, from;
}e[maxn];
int head[maxn], cnt;
queue<int> q;
void add(int u, int v, int w)
{
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].len = w;
    e[cnt].next = head[u];
    head[u] = cnt;
}
int SPFA()
{
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for(int i = head[now]; i ; 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])
                {
                    vis[e[i].to] = 1;
                    q.push(e[i].to);
                }
            }
        }
    }
}
int main()
{
    scanf("%d%d%d",&n,&m,&b);
    for(int i = 1; i <= m; i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
        add(v,u,w);
    }
    for(int i = 1; i <= n; i++)
    dis[i] = inf;
    dis[1] = 0;
    q.push(1);
    vis[1] = 1;
    SPFA();
    for(int i = 1; i <= b; i++)
    {
        int p,q;
        scanf("%d%d",&p,&q);
        printf("%d
",dis[p]+dis[q]);
    }
    return 0;
}