lightoj 1017 - Brush (III)

而为坐标平面上给出n个点，问一把款w的刷子，横着刷k次最多能够刷掉多少个点。

因为是横着刷，所以只考虑纵坐标。

dp[i][k]表示前i个点刷k次最多能够刷的点数。。。然后就是枚举，不过要先处理完所有dp[i][1]的情况。。。

开始想的是dp[i][j][k]表示区间[i,j]刷k次能够刷的最多点，，，但是转移出问题了。。。感觉无法转移状态。。。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 110;
int dp[maxn][maxn];
int A[maxn];
int n,Times,w;
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out1.txt","w",stdout);
	int t,icase = 0;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d%d",&n,&w,&Times);
		for (int i = 1;i <= n;++i)
			scanf("%*d%d",&A[i]);
		sort(A + 1,A +1 + n);
		memset(dp, 0,sizeof dp);
		int ans = 0;
		for (int j = n;j >= 1;--j){
			for (int i = j;i >= 1;--i){
				dp[j][1] += (A[j] - A[i] <= w);
			}
			ans = max(dp[j][1], ans);
		}
		for (int j = 1;j <= n;++j){
			for (int k = 2;k <= Times;++k){
				for (int i = 1;i < j;++i){
					if (A[j] - A[i] > w) dp[j][k] = max(dp[j][k], dp[j][1] + dp[i][k-1]);
				}
				ans = max(ans, dp[j][k]);
			}
		}
		cout << "Case " << ++icase << ": " ;
		cout << ans << endl;
	}
	return 0;
}

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