## Problem

Your are given an undirect connected graph.Every edge has a cost to pass.You should choose a path from S to T and you need to pay for all the edges in your path. However, you can choose at most k edges in the graph and change their costs to zero in the beginning. Please answer the minimal total cost you need to pay.

## 输入描述:

The first line contains five integers n,m,S,T,K.

For each of the following m lines, there are three integers a,b,l, meaning there is an edge that costs l between a and b.

n is the number of nodes and m is the number of edges.

## 输出描述:

An integer meaning the minimal total cost.

## 示例1

``````3 2 1 3 1
1 2 1
2 3 2``````

``1``

## 备注:

1 <= n,m <= 103,1 <= S,T,a, b <= n, 0 <= k <= m,1 <= 106
.
Multiple edges and self loops are allowed.

## 分析

dis[i][j]表示到达i点，已经改了j次边权，此时的最短路。

（1）不用变0技能：转移到dis[dest][j] = dis[now][j] + len

（2）用变0技能：转移到dis[dest][j+1] = dis[now][j]

``````#include<iostream>
#include<vector>
#include<queue>
using namespace std;

const int MAXN = 1e3+10;
const int MAXK = 1e3+10;
const int INF = 0x3f3f3f3f;
struct edge{
int to,cost;
edge(){}
edge(int to,int cost):to(to),cost(cost){}
};
int n,m,S,T,K;
vector<edge>edges[MAXN];

struct node{
int dest,cnt,cost;
node(int dest,int cnt,int cost):dest(dest),cnt(cnt),cost(cost){}

bool operator<(const node &b) const {
return cost>b.cost;
}

};

int dp[MAXN][MAXK];
int dijkstra(int start,int dst){
fill(dp[0],dp[0]+MAXN*MAXK,INF);
priority_queue<node> que;

que.push(node(start,0,0));
dp[start][0] = 0;
while(!que.empty()){
node  temp = que.top(); que.pop();
if(temp.dest == dst) return dp[dst][temp.cnt];
if(dp[temp.dest][temp.cnt] < temp.cost) continue;

for(int i = 0;i<edges[temp.dest].size();i++){
edge e = edges[temp.dest][i];
if(dp[e.to][temp.cnt] > temp.cost + e.cost ){
dp[e.to][temp.cnt] = temp.cost + e.cost;
que.push(node(e.to,temp.cnt,dp[e.to][temp.cnt]));
}

if(dp[e.to][temp.cnt+1] > temp.cost && temp.cnt+1 <=K){
dp[e.to][temp.cnt+1] = temp.cost;
que.push(node(e.to,temp.cnt+1,dp[e.to][temp.cnt+1]));
}
}
}

return INF;

}

int main(){

cin>>n>>m>>S>>T>>K;

int a,b,l;
for(int i=0;i<m;i++){
cin>>a>>b>>l;
edges[a].push_back(edge(b,l));
edges[b].push_back(edge(a,l));
}

int ans = dijkstra(S,T);

printf("%d\n",ans);

return 0;
}``````