lightoj1017 Brush (III) （dp）

1017 - Brush (III)

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

Samir returned home from the contest and got angry after seeing his room dusty. Who likes to see a dusty room after a brain storming programming
contest? After checking a bit he found a brush in his room which has width w. Dusts are defined as 2D points. And since they are scattered everywhere, Samir is a bit confused what to do. He asked Samee and found his idea. So, he attached a
rope with the brush such that it can be moved horizontally (in X axis) with the help of the rope but in straight line. He places it anywhere and moves it. For example, the y co-ordinate of the bottom part of the brush is 2
and its width is 3, so the y coordinate of the upper side of the brush will be 5. And if the brush is moved, all dusts whose y co-ordinates are between 2 and 5 (inclusive) will be cleaned. After cleaning all the dusts in that
part, Samir places the brush in another place and uses the same procedure. He defined a move as placing the brush in a place and cleaning all the dusts in the horizontal zone of the brush.

You can assume that the rope is sufficiently large. Since Samir is too lazy, he doesn't want to clean all the room. Instead of doing it he thought that he would use at most k moves. Now he wants to find the maximum number of dust units he
can clean using at most k moves. Please help him.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a blank line. The next line contains three integers N (1 ≤ N ≤ 100), w (1 ≤ w ≤ 10000) and k (1 ≤ k ≤ 100). N means that there are N dust points. Each of the next N lines
contains two integers: xi yi denoting the coordinates of the dusts. You can assume that (-109 ≤ xi, yi ≤ 109) and all points are distinct.

Output

For each case print the case number and the maximum number of dusts Samir can clean using at most k moves.

Sample Input	Output for Sample Input
2   3 2 1 0 0 20 2 30 2   3 1 1 0 0 20 2 30 2	Case 1: 3 Case 2: 2

 

PROBLEM SETTER: JANE ALAM JAN 

思路：先将点按照纵坐标从小到大排个序，然后再统计当刷子上端刷到 i 这个点的时候前面可以刷到多少个点。

接下来就是状态转移，dp[i][j]表示前i个点刷j次能够刷到的最大值，dp[i][j]=max(dp[i-1][j],dp[i-a[i]][j-1]+a[i])

#include<bits/stdc++.h>
#define N 110

using namespace std;
struct node
{
int x,y;
} p
;

int cmp(node a,node b)
{
return a.y<b.y;
}

int main()
{
int t,n,w,k;
int dp

,a
;
cin>>t;
for(int cas=1; cas<=t; cas++)
{
scanf("%d%d%d",&n,&w,&k);
for(int i=1; i<=n; i++)
scanf("%d%d",&p[i].x,&p[i].y);
sort(p+1,p+1+n,cmp);
memset(a,0,sizeof(a));
for(int i=1; i<=n; i++)
for(int j=i; j>=1; j--)
{
if(p[j].y+w<p[i].y)
break;
a[i]++;
}
//a[i]表示在i处刷一下能覆盖到之前的点的个数
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
for(int j=1; j<=k; j++)
dp[i][j]=max(dp[i-1][j],dp[i-a[i]][j-1]+a[i]);
printf("Case %d: %d\n",cas,dp
[k]);
}
return 0;
}