D. Colorful Points

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本文介绍了一道关于颜色点删除的模拟题，通过算法实现对直线上一系列带有颜色标记的点进行操作，直到无法继续删除。文章详细解释了算法流程，并提供了完整的代码示例。

You are given a set of points on a straight line. Each point has a color assigned to it. For point a, its neighbors are the points which don't have any other points between them and a. Each point has at most two neighbors - one from the left and one from the right.

You perform a sequence of operations on this set of points. In one operation, you delete all points which have a neighbor point of a different color than the point itself. Points are deleted simultaneously, i.e. first you decide which points have to be deleted and then delete them. After that you can perform the next operation etc. If an operation would not delete any points, you can't perform it.

How many operations will you need to perform until the next operation does not have any points to delete?

Input

Input contains a single string of lowercase English letters 'a'-'z'. The letters give the points' colors in the order in which they are arranged on the line: the first letter gives the color of the leftmost point, the second gives the color of the second point from the left etc.

The number of the points is between 1 and 106.

Output

Output one line containing an integer - the number of operations which can be performed on the given set of points until there are no more points to delete.

Examples
inputCopy
aabb
outputCopy
2
inputCopy
aabcaa
outputCopy
1

Note

In the first test case, the first operation will delete two middle points and leave points "ab", which will be deleted with the second operation. There will be no points left to apply the third operation to.

In the second test case, the first operation will delete the four points in the middle, leaving points "aa". None of them have neighbors of other colors, so the second operation can't be applied.

模拟题，我们可以发现第一遍删除后字符串相同的部分会在一起，然后我们把字符串分成多个部分，每个部分由相同的字符组成，中间的部分每次删除的时候减二，两端删除的时候减一，关键一点是合并，在每次删除过后要合并掉相同的部分，这步可能会忘记

#include <bits/stdc++.h>

using namespace  std;

const int inf = 1e6 + 10;
struct{
    char id;
    int shu;
}num[inf], num2[inf];
int main()
{
    string a, b;
    cin >> a;
    bool fou = true;
    int ans = 0;
        b = "";
        int len = a.length();
        for(int i = 0; i < len; i ++){
            fou = true;
             if(i + 1 < len){
                if(a[i + 1] != a[i]){
                    fou = false;
                }
             }
             if(i - 1 >= 0){
                if(a[i - 1] != a[i]) {
                    fou = false;
                }
             }
             if(fou){
                b += a[i];
             }
        }
        if(a == b){
           ans = 0;
        }else {
            ans = 1;
            int len1 = 0;
            len = b.length();
            num[0].shu = 1;
            num[0].id = b[0];
            for(int i = 1; i < len;i ++){
                if(b[i] == b[i - 1]){
                    num[len1].shu ++;
                    num[len1].id = b[i];
                }else {
                    num[++ len1].shu = 1;
                    num[len1].id = b[i];
                }
            }
            len1 ++;//表明有len1个部分
            while(len1 > 1){
                len = len1;
                len1 = 0;
                for(int i = 1; i < len - 1; i ++){
                    num[i].shu -= 2;
                }
                num[0].shu --;
                num[len - 1].shu --;
                for(int i = 0;i < len; i ++){
                    if(num[i].shu > 0){
                        num[len1 ++] = num[i];
                    }
                }
                len = 0;
                num2[0] = num[0];
                for(int i = 1; i < len1; i ++){
                    if(num[i].id == num2[len].id){
                        num2[len].shu += num[i].shu;
                    }else {
                        num2[++ len] = num[i];
                    }
                }
                ans ++;
                for(int i = 0; i <= len; i ++){
                    num[i] = num2[i];
                }
                len1 = len;
                len1 ++;
            }
        }
        cout << ans << endl;
    return 0;
}