划分dp,区间差最小

题目链接：http://codevs.cn/problem/3162/

分析：思路一样，但半天没写出dp方程= =。看了题解才知道，可以先求到 j 个人最多的抄书数目  ，再求最小dp[[i][j]

#include <bits/stdc++.h>
using namespace std;
#define maxn 105
int  m , k;
int dp[maxn][maxn];
int a[maxn];
int s[maxn];
int l[maxn];
int r[maxn];

int main()
{
    cin >> m >> k;
    for(int i = 1 ; i <= m ; ++i)
        scanf("%d",&a[i]),s[i] = s[i-1] + a[i];

    memset(dp,0x7f,sizeof(dp));

    for(int i = 1 ; i <= m ; ++i)
        dp[i][1] = s[i];

    for(int i = 1 ; i <= m ; ++i)
        for(int j = 2 ; j <= k && j <= m ; ++j)
            for(int x = 1 ; x < i ; ++x)
                dp[i][j] = min(dp[i][j],max(dp[x][j-1],s[i]-s[x]));

    int mm = dp[m][k] ;
   // cout << mm << endl;
    int right = m;
    int cnt = 0;
    int sum = 0;
    int i = m;
    while( i >= 1 )
    {
        while( sum + a[i] <= mm && i >= 1 )
        {
            sum += a[i];
            i--;
        }
        l[cnt] = i + 1;
        r[cnt++] = right;
        right = i;
        sum = 0 ;
    }
    for(int i = cnt-1 ; i >= 0 ; --i)
    {
        printf("%d %d\n",l[i],r[i]);
    }
}