NOJ 1072 The longest same color grid（尺取）

Posted 2020-08-15 Blackops

Problem 1072: The longest same color grid

 

Time Limits:  1000 MS   Memory Limits:  65536 KB

64-bit interger IO format:  %lld   Java class name:  Main

 

Description

There are n grid, m kind of color. Grid number 1 to N, color number 1 to M. 
The color of each grid is painted in one of the m colors. Now let you remove the Grid does not exceed K, get a new sequence.
Ask you the same color and continuous length of the grid sequence is the length of the number?

Input

The first line of input T, T group data. (T <= 20)
The next line of input three numbers n, m and k. (1 < n, m，k <= 10^5)
The next line of input n number, indicating that the color of each grid. (1 <= a[i] <= m)

Output

For each group of data output the same color and continuous grid sequence length.

Sample Input

1
10 2 2
1 2 1 2 1 1 2 1 1 2

Output for Sample Input

5

Hint

For example we can delete the fourth position and the seventh position

 

 

题意：给你一个连续的数字染色序列，你最多可以去掉k个格子使得某些格子变成连续的，比如样例的去掉第二和第四个格子，中间的便有5个1是连续的……

学长出的题，刚看到真的是跪了，1e5的范围知道肯定不能两个for去暴力，想过尺取法，然而只是在原序列上进行尺取，这样并不知道到底删除[L，R]中哪几个点才能得到最大的连续长度。

后来学长说是用vector数组记录每一种颜色所出现的下标，对每一种颜色的vector进行尺取，若设左右游标为L、R，则区间内的元素下标就是 $vector[color][Li……Ri]$，区间总长度（闭区间）为 $vector[color][R]-vector[color][L]+1$，所删除的块为记为 $cnt$个，则可获得 $vector[color][R]-vector[color][L]+1-cnt$，依次对每一个颜色的vector进行尺取更新答案即可

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1e5+7;
int arr[N];
vector<int>vec[N];

void init(int m)
{
    for (int i=0; i<=m; ++i)//这里写成 <m 又让我WA了一次，苦逼
        vec[i].clear();
}
int main(void)
{
    int tcase,n,m,k,i,color;
    scanf("%d",&tcase);
    while (tcase--)
    {
        scanf("%d%d%d",&n,&m,&k);
        init(m);
        for (i=1; i<=n; ++i)
        {
            scanf("%d",&color);
            vec[color].push_back(i-1);
            //printf("%d<-%d\n",color,i-1);
        }
        int ans=1;
        for (i=1; i<=m; ++i)
        {
            if(vec[i].empty())
                continue;
            int L=0,R=0;
            int cnt=0;
            int SZ=vec[i].size();
            while (L<SZ)
            {
                while (R<SZ)
                {
                    ++R;
                    if(R>=SZ)
                    {
                        --R;
                        break;
                    }
                    //printf("[%d %d]\n",vec[i][R-1],vec[i][R]);
                    cnt=cnt+vec[i][R]-vec[i][R-1]-1;
                    if(cnt>k)
                    {
                        cnt=cnt-(vec[i][R]-vec[i][R-1]-1);
                        --R;
                        break;
                    }
                }
                int now=vec[i][R]-vec[i][L]-cnt+1;
                if(now>ans)
                    ans=now;
                ++L;
                if(L>=SZ)
                    break;
                else
                    cnt=cnt-(vec[i][L]-vec[i][L-1]-1);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}