# 洛谷P2939 [USACO09FEB]改造路Revamping Trails(最短路)

2018-02-27 08:59:47来源:cnblogs.com作者:自为风月马前卒人点击

## 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

* 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

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

## 输入输出样例

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

`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.

`#include<cstdio>#include<algorithm>#include<queue>#include<cstring>#define Pair pair<int,int>#define F first#define S secondconst int MAXN=1e6+10;using namespace std;inline int read(){    char c=getchar();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}int N,M,K;int dis[MAXN][21],vis[MAXN][21];struct node{    int u,v,w,nxt;}edge[MAXN];int head[MAXN],num=1;inline void AddEdge(int x,int y,int z){    edge[num].u=x;edge[num].v=y;edge[num].w=z;edge[num].nxt=head[x];    head[x]=num++;}void Dijstra(){    memset(dis,0xf,sizeof(dis));    dis[1][0]=0;//第i个点，第j层    priority_queue< pair<int,Pair > > q;    q.push(make_pair(0,make_pair(1,0)));//第一个表示点，第二个表示层     while(q.size()!=0)    {        while(vis[q.top().S.F][q.top().S.S]&&q.size()>0) q.pop();        Pair p=q.top().second;        vis[p.F][p.S]=1;        for(int i=head[p.F];i!=-1;i=edge[i].nxt)        {            int will=edge[i].v;            if(vis[will][p.S]==0&&dis[will][p.S]>dis[p.F][p.S]+edge[i].w)                 dis[will][p.S]=dis[p.F][p.S]+edge[i].w,                q.push(make_pair(-dis[will][p.S],make_pair(will,p.S)));            if(p.S+1<=K&&vis[will][p.S+1]==0&&dis[will][p.S+1]>dis[p.F][p.S])                dis[will][p.S+1]=dis[p.F][p.S],                q.push(make_pair(-dis[will][p.S+1],make_pair(will,p.S+1)));        }    }    printf("%d",dis[N][K]);}int main(){    memset(head,-1,sizeof(head));    N=read();M=read();K=read();    for(int i=1;i<=M;i++)    {        int x=read(),y=read(),z=read();        AddEdge(x,y,z);        AddEdge(y,x,z);    }    Dijstra();    return 0;}`