poj3666-基础DP

题意：然你求如何使得这个序列单调不减或者单调不增然后代价比较小，因为数据比较弱只要求单调不增就可以了

题解：dp[i][j]表示第i个位置改成第j小的数字代价花的最少然后记录一下i-1个位置改成第1-第j小的代价是多少就可以

dp[i][j] = pre[i-1][j]+abs(b[j]-a[i]);

pre[i][j] = min(dp[i][j],pre[i][j-1]);

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mx = 2005;
int dp[mx][mx];
int pre[mx][mx];
int a[mx];
int b[mx];
int main(){
int n;
while(scanf("%d",&n)!=EOF){
memset(dp,inf,sizeof(dp));
memset(pre,inf,sizeof(pre));
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
b[i] = a[i];
pre[0][i] = 0;
}
sort(b+1,b+1+n);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++){
dp[i][j] = pre[i-1][j]+abs(b[j]-a[i]);
pre[i][j] = min(dp[i][j],pre[i][j-1]);
}
int ans = inf;
for(int i = 1; i <= n; i++)
ans = min(ans,dp
[i]);
printf("%d\n",ans);
}
return 0;
}