P3003 [USACO10DEC]蘋果交貨Apple Delivery

題目描述

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.

貝西有兩個又香又脆的紅蘋果要送給她的兩個朋友。當然她可以走的C（1<=C<=200000）條“牛路”都被包含在一種常用的圖中，包含了P（1<=P<=100000）個牧場，分別被標為1..P。沒有“牛路”會從一個牧場又走回它自己。“牛路”是雙向的，每條牛路都會被標上一個距離。最重要的是，每個牧場都可以通向另一個牧場。每條牛路都連接著兩個不同的牧場P1_i和P2_i（1<=P1_i,p2_i<=P)，距離為D_i。所有“牛路”的距離之和不大於2000000000。

現在，貝西要從牧場PB開始給PA_1和PA_2牧場各送一個蘋果（PA_1和PA_2順序可以調換），那麽最短的距離是多少呢？當然，PB、PA_1和PA_2各不相同。

輸入輸出格式

• 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

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

輸入輸出樣例

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 

12 

/*
    基本上是個最短路模板題吧。。
    就是求出t1和t2間的最短路a1
    再求min(dis[s][t1],dis[s][t2])
    兩者相加即可
    本來以為這題沒這麽簡單，隨便一寫結果A了
    我用的堆優化Dij
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 100010
int n,m,s,t1,t2,dis[maxn],num,head[maxn];
bool vis[maxn];
struct node{
    int id,d;
    bool operator < (const node b)const{
        return d>b.d;
    }
};
struct Node{
    int to,pre,v;
}e[200010*2];
void Insert(int from,int to,int v){
    e[++num].to=to;
    e[num].v=v;
    e[num].pre=head[from];
    head[from]=num;
}
void Dij(int st,int tt1,int tt2){
    int flag=0;
    priority_queue<node>q;
    memset(dis,127/3,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[st]=0;node now;
    now.id=st;now.d=0;
    q.push(now);
    while(!q.empty()){
        now=q.top();q.pop();
        int u=now.id;
        if(vis[u])continue;
        vis[u]=1;
        if(u==tt1)flag++;
        if(u==tt2)flag++;
        if(flag==2)return;
        for(int i=head[u];i;i=e[i].pre){
            int to=e[i].to;
            if(dis[u]+e[i].v<dis[to]){
                dis[to]=dis[u]+e[i].v;
                node nxt;nxt.d=dis[to];nxt.id=to;
                q.push(nxt);
            }
        }
    }
}
int main(){
    //freopen("Cola.txt","r",stdin);
    scanf("%d%d%d%d%d",&m,&n,&s,&t1,&t2);
    int x,y,z;
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&x,&y,&z);
        Insert(x,y,z);
        Insert(y,x,z);
    }
    int a1,a2=0x7fffffff;
    Dij(s,t1,t2);
    a1=min(dis[t1],dis[t2]);
    Dij(t1,t2,t2);
    a2=min(dis[t2],a2);
    Dij(t2,t1,t1);
    a2=min(dis[t1],a2);
    cout<<a1+a2;
}

洛谷P3003 [USACO10DEC]蘋果交貨Apple Delivery