xtu summer individual 6 E - Find Metal Mineral
Find Metal Mineral
This problem will be judged on HDU. Original ID: 4003
64-bit integer IO format: %I64d Java class name: Main
Input
In each case:
The first line specifies three integers N, S, K specifying the numbers of metal mineral, landing site and the number of robots.
The next n‐1 lines will give three integers x, y, w in each line specifying there is a path connected point x and y which should cost w.
1<=N<=10000, 1<=S<=N, 1<=k<=10, 1<=x, y<=N, 1<=w<=10000.
Output
Sample Input
3 1 1
1 2 1
1 3 1
3 1 2
1 2 1
1 3 1
Sample Output
3
2
In the first case: 1->2->1->3 the cost is 3; In the second case: 1->2; 1->3 the cost is 2;
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
struct arc {
int to,w;
};
vector<arc>g[];
int dp[][],n,s,k;
void dfs(int u,int fa){
for(int i = ; i < g[u].size(); i++){
int v = g[u][i].to;
if(v == fa) continue;
dfs(v,u);
for(int t = k; t >= ; t--){
dp[u][t] += dp[v][] + *g[u][i].w;
for(int j = ; j <= t; j++)
dp[u][t] = min(dp[u][t],dp[u][t-j]+dp[v][j]+j*g[u][i].w);
}
}
}
int main(){
while(~scanf("%d%d%d",&n,&s,&k)){
for(int i = ; i <= n; i++)
g[i].clear();
for(int i = ; i < n; i++){
int u,v,w;
scanf("%d %d %d",&u,&v,&w);
g[u].push_back((arc){v,w});
g[v].push_back((arc){u,w});
}
memset(dp,,sizeof(dp));
dfs(s,-);
printf("%d\n",dp[s][k]);
}
return ;
}
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