BZOJ 1579: [Usaco2009 Feb]Revamping Trails 道路升级( 最短路 )

最短路...多加一维表示更新了多少条路

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#include<cstdio>

#include<algorithm>

#include<queue>

#include<cstring>

#include<iostream>

 

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

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

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 10000 + 5;

const int maxm = 50000 + 5;

const int maxk = 20 + 5;

const ll INF = 1LL << 50;

 

struct edge {

int to , dist;

edge* next;

};

 

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

edge* head[ maxn ];

 

void init() {

pt = EDGE;

clr( head , 0 );

}

 

inline void add( int u , int v , ll 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 )

 

ll d[ maxn ][ maxk ];

 

int n , K;

 

struct Node {

int x , k;

ll d;

bool operator < ( const Node &o ) const {

return d > o.d;

}

};

 

priority_queue< Node > Q;

 

void dijkstra() {

rep( i , n ) {

   if( i ) 

   

   rep( j , K ) d[ i ][ j ] = INF;

   

else 

   rep( j , K )

       d[ 0 ][ i ] = 0 , Q.push( ( Node ) { 0 , j , 0 } );

       

}

while( ! Q.empty() ) {

Node o = Q.top();

Q.pop();

int x = o.x , k = o.k;

ll dist = o.d;

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

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

int to = e -> to;

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

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

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

}

if( k + 1 < K && d[ to ][ k + 1 ] > dist ) {

d[ to ][ k + 1 ] = dist;

Q.push( ( Node ) { to , k + 1 , dist } );

}

}

}

}

int main() {

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

init();

int m;

cin >> n >> m >> K;

K++;

while( m-- ) {

int u , v , d;

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

u-- , v--;

add_edge( u , v , d );

}

dijkstra();

ll ans = INF;

rep( i , K )

   ans = min( ans , d[ n - 1 ][ i ] );

   

cout << ans << "\n";

return 0;

}

 

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1579: [Usaco2009 Feb]Revamping Trails 道路升级

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 1378  Solved: 374
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Description

每天,农夫John需要经过一些道路去检查牛棚N里面的牛. 农场上有M(1<=M<=50,000)条双向泥土道路,编号为1..M. 道路i连接牛棚P1_i和P2_i (1 <= P1_i <= N; 1 <= P2_i<= N). John需要T_i (1 <= T_i <= 1,000,000)时间单位用道路i从P1_i走到P2_i或者从P2_i 走到P１_i 他想更新一些路经来减少每天花在路上的时间.具体地说,他想更新K (1 <= K <= 20)条路经，将它们所须时间减为０．帮助FJ选择哪些路经需要更新使得从１到N的时间尽量少.

Input

* 第一行: 三个空格分开的数: N, M, 和 K * 第2..M+1行: 第i+1行有三个空格分开的数：P1_i, P2_i, 和 T_i

Output

* 第一行: 更新最多Ｋ条路经后的最短路经长度．

Sample Input

4 4 1
1 2 10
2 4 10
1 3 1
3 4 100

Sample Output

1

HINT

K是1; 更新道路3->4使得从３到４的时间由１００减少到０. 最新最短路经是1->3->4,总用时为１单位. N<=10000

Source

Gold