[2016-03-28][POJ][3666][]Making the Grade]

• 时间:2016-03-28 17:23:08 星期一

• 题目编号:[2016-03-28][POJ][3666][]Making the Grade]

• 分析:dp[i][j]表示把改到第i个数,且把a[i]改成b[i]需要的最少代价,已知b[j]递增,dp[i][j]由dp[i - 1][k]递推过来,如果改成非增,那么就要求 a[j] <= a[k],所以dp[i][j] = min(dp[i - 1][k] + d) k < j;

#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn =  2000 + 10;
int a[maxn],b[maxn],n;
int dp[maxn][maxn];
bool cmp(int x,int y){
        return x > y;
}
int main(){
        scanf("%d",&n);
        for(int i = 0;i < n ; ++i){
                scanf("%d",&a[i]);
        }
        memcpy(b,a,sizeof(int) * (n + 1));
        sort(b,b + n);
        memset(dp,0x7f,sizeof(dp));
        for(int i = 0;i < n ; ++i){
                dp[0][i] = abs(a[0] - b[i]);
        }
        for(int i = 1;i < n ; ++i){
                int mindp = 0x7f7f7f7f;
                for(int j = 0;j < n;++j){
                        mindp = min(dp[i-1][j],mindp);
                        dp[i][j] = min(dp[i][j] , mindp + abs(a[i] - b[j]));
                }
        }
        int ans = 0x7f7f7f7f;
        for(int i = 0;i < n ; ++i){
                ans = min(dp[n - 1][i] ,ans);
        }
        sort(b,b + n,cmp);
        memset(dp,0x7f,sizeof(dp));
        for(int i = 0;i < n ; ++i){
                dp[0][i] = abs(a[0] - b[i]);
        }
        for(int i = 1;i < n ; ++i){
                int mindp = 0x7f7f7f7f;
                for(int j = 0;j < n;++j){
                        mindp = min(dp[i-1][j],mindp);
                        dp[i][j] = min(dp[i][j] , mindp + abs(a[i] - b[j]));
                }
        }
        for(int i = 0;i < n ; ++i){
                ans = min(dp[n - 1][i] ,ans);
        }
        printf("%d\n",ans);       
        return 0;
}

[2016-03-28][POJ][3666][]Making the Grade]

原文：http://www.cnblogs.com/qhy285571052/p/0f251142b5970f12318ac5ce3612fb50.html