POJ-3666 Making the Grade

题目传送门

题解： B[i]的值肯定是A[i]的值的某一个，将A[i]sort一下 F[i][j]表示 处理道第i个A[i]    以sort完后的新数组来说 B[j]结尾的最小花费。

代码：

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 2e3 + 100;
int a[N], b[N];
LL F[N][N];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]), b[i] = a[i];
    sort(b+1, b+1+n);

    for(int i = 1; i <= n; i++){
        LL mn = INF;
        for(int j = 1; j <= n; j++) F[i][j] = INF;
        for(int j = 1; j <= n; j++){
            mn = min(mn, F[i-1][j]);
            F[i][j] = min(F[i][j], mn + abs(a[i]-b[j]));
        }
    }
    LL ans = INF;
    for(int j = 1; j <= n; j++) ans = min(F[n][j], ans);
    for(int i = 1; i <= n; i++){
        LL mn = INF;
        for(int j = 1; j <= n; j++) F[i][j] = INF;
        for(int j = n; j >= 1; j--){
            mn = min(mn, F[i-1][j]);
            F[i][j] = min(F[i][j], mn + abs(a[i]-b[j]));
        }
    }
    for(int j = 1; j <= n; j++) ans = min(F[n][j], ans);
    printf("%lld
", ans);
    return 0;
}