Making the Grade(POJ-3666)
作者:互联网
题目大意是给出一个长度为 n 的序列,要求使序列变为单调上升或单调不减序列(非严格),问花费的最少代价?
转移方程是:dp[i][j]表示前i个元素的最后一个元素为全部元素第j小时的最小代价
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<vector>
#include<map>
#define PI acos(-1.0)
#define LL long long
#define INF 0x3f3f3f3f
const int maxn=2000+5;
const int MOD=10007;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,1,-1};
using namespace std;
int a[maxn];
int b[maxn];
int dp[maxn][maxn];
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++) {
cin>>a[i];
b[i]=a[i];
}
sort(b+1,b+1+n);
for(int i=1;i<=n;i++){
int minn=INF;
for(int j=1;j<=n;j++){
minn=min(minn,dp[i-1][j]);
dp[i][j]=abs(a[i]-b[j])+minn;
}
}
int res=INF;
for(int i=1;i<=n;i++) res=min(res,dp[n][i]);
cout<<res<<endl;
}
标签:3666,const,minn,Grade,int,maxn,Making,include,dp 来源: https://blog.csdn.net/hahadelaochao/article/details/115588894