BZOJ 2100: [Usaco2010 Dec]Apple Delivery( 最短路 )

跑两遍最短路就好了.. 

话说这翻译2333 

----------------------------------------------------------------------

#include<cstdio>

#include<queue>

#include<algorithm>

#include<cstring>

#include<iostream>

 

#define rep( i , n ) for( int i = 0 ; i < n ; ++i )

#define Rep( i , n ) for( int i = 1 ; i <= n ; ++i )

#define clr( x , c ) memset( x , c , sizeof( x ) )

 

using namespace std;

 

const int maxn = 100000 + 5;

const int maxm = 200000 + 5;

const int inf = int( 2e9 ) + 1;

 

struct edge{

int to , dist;

edge* next;

};

 

edge* pt , EDGE[ maxm << 1 ];

edge* head[ maxn ];

 

void init() {

pt = EDGE;

clr( head , 0 );

}

 

void add( int u , int v , int d ) {

pt -> to = v;

pt -> dist = d;

pt -> next = head[ u ];

head[ u ] = pt++;

}

 

#define add_edge( u , v , d ) add( u , v , d ) , add( v , u , d )

 

int d[ maxn ];

 

struct Node {

int x , d;

bool operator < ( const Node &rhs ) const {

return d > rhs.d;

}

};

 

priority_queue< Node > Q;

 

void dijkstra( int S , int N ) {

rep( i , N ) d[ i ] = inf;

d[ S ] = 0;

Q.push( ( Node ) { S , 0 } );

while( ! Q.empty() ) {

Node o = Q.top();

Q.pop();

int x = o.x , dist = o.d;

if( dist != d[ x ]) continue;

for( edge* e = head[ x ] ; e ; e = e -> next ) {

int to = e -> to;

if( d[ to ] > dist + e -> dist ) {

d[ to ] = dist + e -> dist;

Q.push( ( Node ) { to , d[ to ] } );

}

}

}

}

 

int main() {

freopen( "test.in" , "r" , stdin );

init();

int m , n , S , a , b;

cin >> m >> n >> S >> a >> b;

S-- , a-- , b--;

while( m-- ) {

int u , v , d;

scanf( "%d%d%d" , &u , &v , &d );

u-- , v--;

add_edge( u , v , d );

}

dijkstra( S , n );

int dist[ 2 ] = { d[ a ] , d[ b ] };

dijkstra( a , n );

int Dist = d[ b ];

cout << min( dist[ 0 ] , dist[ 1 ] ) + Dist << " ";

return 0;

}

---------------------------------------------------------------------- 

2100: [Usaco2010 Dec]Apple Delivery

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 479  Solved: 169
[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.........

Source

Silver