hdu1421搬寝室DP（ACM题目，动态规划）

hdu1421搬寝室DP（ACM题目，动态规划）

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学习总结 数据结构和算法
2011-11-07 19:14
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分析：dp[i][j]表示前i个元素中 j 对数的最小差平方和

dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + (weight[i] - weight[i-1])*(weight[i] - weight[i-1]));

code:

#include<stdio.h>

#include<stdlib.h>

#define min(a,b) ((a) > (b) ? (b) : (a))

#define MaxAllow 2147483648

const int num = 2001;

int cmp(const void* a, const void *b)

{

return *(int*)a - *(int*)b;

}

int dp[num][1001]; //数组要在函数外定义，不然太大不行，无语，又卡了n久

int main()

{

int n, k;

int i, j;

int weight[num];

while(scanf("%d%d", &n, &k) != EOF)

{

for(i = 1; i <= n; i++)

scanf("%d", &weight[i]);

qsort(weight+1, n, sizeof(weight[0]), cmp);

for(i = 0; i <= n; i++)

for(j = 1; j <= k; j++)

dp[i][j] = MaxAllow;

dp[0][0] = 0;

for(i = 2; i <= n; i++)

for(j = 1; j*2 <= i; j++)

{

dp[i][j] = min(dp[i-1][j],dp[i-2][j-1]+(weight[i]-weight[i-1])*(weight[i] - weight[i-1]));

}

printf("%d\n", dp
[k]);

}

return 0;

}