P2939 [USACO09FEB]改造路Revamping Trails【题解】

题意翻译

约翰一共有$N$个牧场.由$M$条布满尘埃的小径连接.小径可 以双向通行.每天早上约翰从牧场$1$出发到牧场$N$去给奶牛检查身体.

通过每条小径都需要消耗一定的时间.约翰打算升级其中$K$条小径，使之成为高速公路.在高速公路上的通行几乎是瞬间完成的，所以高速公路的通行时间为$0$.

请帮助约翰决定对哪些小径进行升级，使他每天从$1$号牧场到第$N$号牧场所花的时间最短

题目描述

Farmer John dutifully checks on the cows every day. He traverses some of the M (1 <= M <= 50,000) trails conveniently numbered 1…M from pasture 1 all the way out to pasture N (a journey which is always possible for trail maps given in the test data). The N (1 <= N <= 10,000) pastures conveniently numbered 1…N on Farmer John’s farm are currently connected by bidirectional dirt trails. Each trail i connects pastures P1_i and P2_i (1 <= P1_i <= N; 1 <= P2_i <= N) and requires T_i (1 <= T_i <= 1,000,000) units of time to traverse.

He wants to revamp some of the trails on his farm to save time on his long journey. Specifically, he will choose K (1 <= K <= 20) trails to turn into highways, which will effectively reduce the trail’s traversal time to 0. Help FJ decide which trails to revamp to minimize the resulting time of getting from pasture 1 to N.

TIME LIMIT: 2 seconds

输入输出格式

输入格式：

• Line 1: Three space-separated integers: N, M, and K

• Lines 2…M+1: Line i+1 describes trail i with three space-separated integers: P1_i, P2_i, and T_i

输出格式：

• Line 1: The length of the shortest path after revamping no more than K edges

输入输出样例

输入样例#1：

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

输出样例#1： 复制

1

说明

K is 1; revamp trail 3->4 to take time 0 instead of 100. The new shortest path is 1->3->4, total traversal time now 1.

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这道题是分层图最短路问题，顾名思义，就是在分层图上跑最短路

这样的题好像都挺裸的

刚刚写了一篇类似的，不懂可以看这个Luogu P4568 [JLOI2011]飞行路线【题解】

这两个题除了输入输出就可以说是一样的。

直接上代码吧。

代码：

#include<iostream>
#include<cstdio>
#include<ctype.h>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
typedef pair<int,int> pairs;
priority_queue<pairs,vector<pairs>,greater<pairs> > q;
inline int read(){
	int x=0,f=0;char ch=getchar();
	while(!isdigit(ch))f|=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+(ch^48),ch=getchar();
	return f?-x:x;
}
int head[220007],cnt;
bool vis[220007];
int dis[220007];
int n,m,k,s=1,ans=99999999;
struct Edge{
	int next,to,w;
}edge[40000007];
inline void add_edge(int from,int to,int w){
	edge[++cnt].next=head[from];edge[cnt].w=w;
	edge[cnt].to=to;head[from]=cnt;
}
inline void dijkstra(){
	memset(dis,0x3f,sizeof dis);
	dis[s]=0;q.push(make_pair(dis[s],s));
	while(!q.empty()){
		int x=q.top().second;q.pop();
		if(vis[x])continue;vis[x]=1;
		for(int i=head[x];i;i=edge[i].next){
			int to=edge[i].to;
			if(dis[to]>dis[x]+edge[i].w){
				dis[to]=dis[x]+edge[i].w;
				q.push(make_pair(dis[to],to));
			}
		}
	}
}
int main(){
	n=read(),m=read(),k=read();
	for(int i=1;i<=m;++i){
		int u=read(),v=read(),w=read();
		add_edge(u,v+n,0),add_edge(v,u+n,0);
		for(int j=0;j<k;++j)add_edge(u+j*n,v+j*n+n,0),add_edge(v+j*n,u+j*n+n,0);
		for(int j=0;j<=k;++j){
			add_edge(u+j*n,v+j*n,w);
			add_edge(v+j*n,u+j*n,w);
		}
	}
	dijkstra();
	for(int i=1;i<=k+1;++i)ans=min(ans,dis[i*n]);
	printf("%d",ans);
	return 0;
}