codeforces1013E - DP

Hills

题目大意

从n个数选取k个数，如果选第i个数，必须h[i−1]<h[i]>h[i+1]$h[i-1]<h[i]>h[i+1]$,每花费1可以将对应的h减小1，问你选取k个数的最小花费。输出k=1,2,3,⋯(n+1)/2$1,2,3,\cdots (n+1)/2$的这些情况。

数据范围

1≤n≤5000,1≤hi≤100000$1\le n\le 5000,1\le h_i\le 100000$

解题思路

dp[i][j][k]$dp[i][j][k]$代表前i个数选j个状态为k的最小花费，k取0和1，dp[i][j][0]$dp[i][j][0]$代表前i个数选j个最后一个数是前i个的最小花费，dp[i][j][1]$dp[i][j][1]$代表前i个数选j个最后一个数是i的最小花费。

那么dp[i][j][0]=min(dp[i−1][j][0],dp[i−1][j][1]+max(0,h[i]−h[i−1]+1))$dp[i][j][0]=min(dp[i-1][j][0],dp[i-1][j][1]+max(0,h[i]-h[i-1]+1))$

同样dp[i][j][1]=min(dp[i−2][j−1][0]+max(0,h[i−1]−h[i]+1,dp[i−2][j−1][1]+max(0,h[i−1]−min(h[i−2],h[i]))))$dp[i][j][1]=min(dp[i-2][j-1][0]+max(0,h[i-1]-h[i]+1,dp[i-2][j-1][1]+max(0,h[i-1]-min(h[i-2],h[i]))))$

AC代码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int INF = 1e9;
const int maxn = 5000;
int dp[maxn + 5][maxn / 2 + 5][2];
//dp[i][j][0]代表前i个选j个且以i之前的为最后一个的最小花费；
//dp[i][j][1]代表前i个选j个且以i为最后一个的最小花费；
int h[maxn + 5];
int n;
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)scanf("%d", &h[i]);
    int m = (n + 1) / 2;
    for(int i = 0; i <= n; i++) {
        for(int j = 0; j <= m; j++)
            dp[i][j][0] = dp[i][j][1] = INF;
            dp[i][0][0] = 0;
    }
    dp[1][1][1] = 0;
    for(int i = 2; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            dp[i][j][0] = min(dp[i - 1][j][0], dp[i - 1][j][1] + max(0, h[i] - h[i - 1] + 1));
            dp[i][j][1] = min(dp[i - 2][j - 1][1] + max(0, h[i - 1] - min(h[i - 2], h[i]) + 1), dp[i - 2][j - 1][0] + max(0, h[i - 1] - h[i] + 1));
        }
    for(int i = 1; i <= m; i++)printf("%d ", min(dp[n][i][0], dp[n][i][1]));
    return 0;
}