搬寝室
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25567 Accepted Submission(s): 8783
题意:要从n件物品中挑走k对物品(2*k<=n),问消耗的最小体力是多少。
解题思路:将物品的重量按照小-》大排序,dp[i][j]表示前 i 个物品选取 j 对,如果第 i 个物品不选,那么dp[i][j]=dp[i-1][j],如果选取第 i 件物品,那么dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+pow(a[i]-a[i-1],2).
<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int INF=INT_MAX;
int dp[2005][1005];
int a[20005];
int computer(int i)
{
return ((a[i]-a[i-1])*(a[i]-a[i-1]));
}
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
sort(a+1,a+n+1);
for(int i=0;i<=n;i++){
for(int j=1;j<=k;j++)
dp[i][j]=INF;
}
dp[0][0]=0;
for(int i=2;i<=n;i++){
for(int j=1;j+j<=i;j++){
dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+computer(i));
}
}
printf("%d\n",dp[n][k]);
}
return 0;
}
</span>