BZOJ 2100 Usaco2010 Dec Apple Delivery _USACO

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2100: [Usaco2010 Dec]Apple Delivery Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 894 Solved: 348
[Submit][Status][Discuss] DescriptionBessie 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:

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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.
一张P个点的无向图，C条正权路。
CLJ要从Pb点（家）出发，既要去Pa1点NOI赛场拿金牌，也要去Pa2点CMO赛场拿金牌。（途中不必回家）
可以先去NOI，也可以先去CMO。
当然神犇CLJ肯定会使总路程最小，输出最小值。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 Input 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

Sample Output 12
HINT
求翻译.........站内PM我吧.........

SourceSilver

因为图是无向图，所以做两次spfa，分别以pa1和pa2为起点即可

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

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