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
题意翻译
约翰一共有 NN 个牧场.由 MM 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 11 出发到牧场 NN 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 KK 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 00。
请帮助约翰决定对哪些小径进行升级，使他每天从 11 号牧场到第 NN 号牧场所花的时间最短。
输入输出样例
输入 #14 4 1 1 2 10 2 4 10 1 3 1 3 4 100输出 #1
1
说明/提示
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.
思路：分层图，将链接两层的边权值设为零，共建k+1层，这时你通过的连接两层的路共有k条
最后在这个图上跑一边最短路，遍历每个终点选出最小值
#include<bits/stdc++.h>
using namespace std;
struct edge{
int v,w,ne;
}e[20000010];
struct node{
int d,dis;
bool operator <(const node x)const{
return dis>x.dis;
}
};
int n,m,k,cnt = 0,h[4000010],dist[2000010];
priority_queue<node>q;
void add(int u,int v,int w){
cnt++;
e[cnt].ne = h[u];
e[cnt].v = v;
e[cnt].w = w;
h[u] = cnt;
}
void dij(int s){
memset(dist,0x3f,sizeof(dist));
dist[s] = 0;
q.push((node){s,0});
while (q.size()){
node t = q.top();
q.pop();
int u = t.d,dis = t.dis;
if (dis!=dist[u]) continue;
for(int i = h[u];i;i = e[i].ne)
if(dist[e[i].v]>dist[u]+e[i].w){
dist[e[i].v] = dist[u]+e[i].w;
q.push((node){e[i].v,dist[e[i].v]});
}
}
}
int main(){
scanf("%d%d%d",&n,&m,&k);
for(int i = 1;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);add(v,u,w);
for(int j = 1;j<=k;j++){
add(n*j+u,n*j+v,w);
add(n*j+v,n*j+u,w);
add(n*(j-1)+u,n*j+v,0);
add(n*(j-1)+v,n*j+u,0);
}
}
dij(1);
int ans = dist[n];
for (int i = 1;i<=k;i++)
ans = min(ans,dist[i*n+n]);
printf("%d",ans);
return 0;
}
综上所述，我还是太菜了
2022-10-28 21:43:30
标签：node,cnt,dist,int,Trails,小径,Revamping,P2939,dis From： https://www.cnblogs.com/cztq/p/16837622.html