BZOJ 2100 Usaco2010 Dec Apple Delivery

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2100: [Usaco2010 Dec]Apple DeliveryTime Limit: 10 Sec  Memory Limit: 64 MB
Submit: 894  Solved: 348
[Submit][Status][Discuss]Description

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:  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.

一張P個點的無向圖，C條正權路。
CLJ要從Pb點（家）出發，既要去Pa1點NOI賽場拿金牌，也要去Pa2點CMO賽場拿金牌。（途中不必回家）
可以先去NOI，也可以先去CMO。
當然神犇CLJ肯定會使總路程最小，輸出最小值。Input

* 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

Output

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

Sample Input9 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



Sample Output12
HINT

 

求翻譯.........站內PM我吧.........

 

Source

Silver

 

因為圖是無向圖，所以做兩次spfa，分別以pa1和pa2為起點即可

 1 #include <bits/stdc++.h> 2 #define ll long long 3 #define inf 1000000 4 #define eps 1e-7 5 using namespace std; 6 inline int read(){ 7     int x=0;int f=1;char ch=getchar(); 8     while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();} 9     while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}10     return x*f;11 }12 const int MAXN=1e6+10;13 struct node{14     int y,next,v;15 }e[MAXN];16 int linkk[MAXN],len,n,m,dis[MAXN],vis[MAXN];17 inline void insert(int xx,int yy,int vv){18     e[++len].y=yy;e[len].next=linkk[xx];e[len].v=vv;linkk[xx]=len;19 }20 void spfa(int st){21     deque < int > q;22     memset(dis,10,sizeof(dis));23     memset(vis,0,sizeof(vis));24     q.push_back(st);dis[st]=0;vis[st]=1;25     while(!q.empty()){26         int tn=q.front();q.pop_front();vis[tn]=0;27         for(int i=linkk[tn];i;i=e[i].next){28             if(dis[e[i].y]>dis[tn]+e[i].v){29                 dis[e[i].y]=dis[tn]+e[i].v;30                 if(!vis[e[i].y]){31                     vis[e[i].y]=1;32                     if(!q.empty()&&dis[q.front()]>dis[e[i].y]) q.push_front(e[i].y);33                     else q.push_back(e[i].y);34                 }35             }36         }37     }38 }39 int main(){40     m=read();n=read();int s=read();int t1=read();int t2=read();41     for(int i=1;i<=m;i++){42         int xx=read();int yy=read();int vv=read();;43         insert(xx,yy,vv);insert(yy,xx,vv);44     }45     spfa(t1);46     int ans=dis[t2]+dis[s];47     spfa(t2);48     ans=min(ans,dis[t1]+dis[s]);49     cout<<ans<<endl;50     return 0;51 }

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BZOJ 2100 Usaco2010 Dec Apple Delivery