USACO09FEB改造路Revamping Trails

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  • 题意
    有$N (1 \le N \le 10000)$个点，$M(1 \le M \le 50000)$条边的无向图。可以使$K$条边的边权变成0，求点$1$到点$N$的最短距离。

  • Solution

    • 设$dis[i][j]$代表到点$i$改造了$j$条路的最短路。
    • 数据卡spfa

  • tips

    • 改成Dijkstra竟然还WA了一次，说明好久没写生疏了。
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#include <bits/stdc++.h>
using namespace std ;
void Read ( int &x, char c = getchar() ) {
	for ( x = 0 ; !isdigit(c) ; c = getchar() )  ;
	for ( ; isdigit(c) ; c = getchar() ) x = 10*x + c - '0' ;
}
const int maxn = 1e4+5, maxm = 1e6+5, zhf = 0x3f3f3f3f ;
int n, m, cnt, e, Begin[maxn], Next[maxm], To[maxm], W[maxm] ;
void add ( int x, int y, int z ) {
	To[++e] = y ;
	Next[e] = Begin[x] ;
	Begin[x] = e ;
	W[e] = z ;
}
int dis[maxn][25] ;
bool inq[maxn][25] ;
struct node {
	int x, stp, dis ;
	friend bool operator < ( node a, node b ) {
		return a.dis > b.dis ;
	}
} ;
priority_queue <node> Q ;
void Dijkstra( int x ) {
	int i, j, u ;
	for ( i = 1 ; i <= n ; i ++ )
		for ( j = 0 ; j <= cnt ; j ++ )
		dis[i][j] = zhf ;
	dis[x][0] = 0 ;
	Q.push( (node){x, 0, 0} ) ;
	node t ;
	while ( !Q.empty() ) {
		t = Q.top() ;
		Q.pop() ;
		x = t.x ;
		if ( inq[x][t.stp] ) continue ;
		inq[x][t.stp] = 1 ;
		for ( i = Begin[x] ; i ; i = Next[i] ) {
			u = To[i] ;
			for ( j = 0 ; j <= cnt ; j ++ ) {
				if ( dis[u][j] > dis[x][j] + W[i] ) {
					dis[u][j] = dis[x][j] + W[i] ;
					Q.push( (node){u, j, dis[u][j]} ) ;
				}
				if ( j < cnt && dis[u][j+1] > dis[x][j] ) {
					dis[u][j+1] = dis[x][j] ;
					Q.push( (node){u, j+1, dis[u][j+1]} ) ;
				}
			}
		}
	}
	printf ( "%d\n", dis[n][cnt] ) ;
}
int main() {
	int i, x, y, z ;
	Read(n) ; Read(m) ; Read(cnt) ;
	for ( i = 1 ; i <= m ; i ++ ) {
		Read(x) ; Read(y) ; Read(z) ;
		add ( x, y, z ) ;
		add ( y, x, z ) ;
	}
	Dijkstra(1) ;
	return 0 ;
}