【洛谷 3003】[USACO10DEC]Apple Delivery S

题目描述

Bessie has two crisp red apples to deliver to two of her friends in the herd. Of course, she travels the C (1 <= C <= 200,000)

cowpaths which are arranged as the usual graph which connects P (1 <= P <= 100,000) pastures conveniently numbered from 1..P: no cowpath leads from a pasture to itself, cowpaths are bidirectional, each cowpath has an associated distance, and, best of all, it is always possible to get from any pasture to any other pasture. Each cowpath connects two differing pastures P1_i (1 <= P1_i <= P) and P2_i (1 <= P2_i <= P) with a distance between them of D_i. The sum of all the distances D_i does not exceed 2,000,000,000.

What is the minimum total distance Bessie must travel to deliver both apples by starting at pasture PB (1 <= PB <= P) and visiting pastures PA1 (1 <= PA1 <= P) and PA2 (1 <= PA2 <= P) in any order. All three of these pastures are distinct, of course.

Consider this map of bracketed pasture numbers and cowpaths with distances:

               3        2       2
           [1]-----[2]------[3]-----[4]
             \     / \              /
             7\   /4  \3           /2
               \ /     \          /
               [5]-----[6]------[7]
                    1       2

If Bessie starts at pasture [5] and delivers apples to pastures [1] and [4], her best path is:

5 -> 6-> 7 -> 4* -> 3 -> 2 -> 1*

with a total distance of 12.

输入格式

* Line 1: Line 1 contains five space-separated integers: C, P, PB, PA1, and PA2

* Lines 2..C+1: Line i+1 describes cowpath i by naming two pastures it connects and the distance between them: P1_i, P2_i, D_i

输出格式

* Line 1: The shortest distance Bessie must travel to deliver both apples

题意翻译

贝西有两个又香又脆的红苹果要送给她的两个朋友。当然她可以走的CC（1<=C<=2000001<=C<=200000）条“牛路”都被包含在一种常用的图中，包含了PP（1<=P<=1000001<=P<=100000）个牧场，分别被标为1..P1..P。没有“牛路”会从一个牧场又走回它自己。“牛路”是双向的，每条牛路都会被标上一个距离。最重要的是，每个牧场都可以通向另一个牧场。每条牛路都连接着两个不同的牧场P1\_iP1_i和P2\_iP2_i（1<=P1\_i,p2\_i<=P1<=P1_i,p2_i<=P)，距离为D\_iD_i。所有“牛路”的距离之和不大于20000000002000000000。

现在，贝西要从牧场PBPB开始给PA\_1PA_1和PA\_2PA_2牧场各送一个苹果（PA\_1PA_1和PA\_2PA_2顺序可以调换），那么最短的距离是多少呢？当然，PBPB、PA\_1PA_1和PA\_2PA_2各不相同。

输入输出样例

输入 #1复制

9 7 5 1 4 
5 1 7 
6 7 2 
4 7 2 
5 6 1 
5 2 4 
4 3 2 
1 2 3 
3 2 2 
2 6 3 

输出 #1复制

12 

题解：题目卡SPFA，因此进行stf优化即可。我也是才学的（其实我觉得背下来就ojbk？）
     注意dis[v]<dis[q.front()] q.front()不能写u，我就是这里想了好久哦哈哈哈

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int oo=0x3f3f3f3f;
const int N=400002;

int m,n,S,T1,T2,head[N];
int cnt,vis[N],dis[N],x,y,z;

struct node{
    int next;
    int to;
    int w;
}e[N];

void add(int x,int y,int z){
    e[++cnt].to=y;
    e[cnt].w=z;
    e[cnt].next=head[x];
    head[x]=cnt;
}

int spfa(int s,int t){
    //queue<int>q;
    std::deque<int>q;
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    vis[s]=1; dis[s]=0; q.push_front(s);
    while(q.size()){
        int u=q.front(); q.pop_front(); vis[u]=0;
        for(int i=head[u];i;i=e[i].next){
            int v=e[i].to;
            if(dis[v]>dis[u]+e[i].w){
                dis[v]=dis[u]+e[i].w;
                if(!vis[v]){
                    vis[v]=1; //q.push_front(v); 
                    if(q.size() && dis[v]<dis[q.front()])
                       q.push_front(v);
                    else q.push_back(v);
                }
            }
        }
    } 
    return dis[t];
}
int ans=-1;
int main(){
    freopen("3033.in","r",stdin);
    freopen("3033.out","w",stdout);
    scanf("%d %d %d %d %d",&m,&n,&S,&T1,&T2);
    for(int i=1;i<=m;i++){
        scanf("%d %d %d",&x,&y,&z);
        add(x,y,z); add(y,x,z);
    }
    cout<<std::min(spfa(S,T1),spfa(S,T2))+spfa(T1,T2);
    return 0;
}