CodeForces 909D

Posted 2020-11-21 tiberius

题意略。

思路：

将字符分桶，然后暴力去扫，扫完合并。假设有k个桶，每个桶里有n / k个数，那么我们应该要扫 n / (2 * k)次，每次的复杂度是k，最后算得复杂度是O(n)。

详见代码：

#include<bits/stdc++.h>
#define maxn 1000005
using namespace std;

struct node{
    int color;
    int numb;
    node(int a = 0,int b = 0){
        color = a,numb = b;
    }
};

node store[maxn];
int tail;
char str[maxn];

int main(){
    scanf("%s",str);
    store[tail++] = node(str[0] - ‘a‘,1);
    for(int i = 1;str[i];++i){
        if(store[tail - 1].color == str[i] - ‘a‘) store[tail - 1].numb += 1;
        else store[tail++] = node(str[i] - ‘a‘,1);
    }
    int ans = 0;
    while(tail > 1){
        for(int i = 0;i < tail;++i){
            if(i == 0 || i == tail - 1) store[i].numb -= 1;
            else store[i].numb -= 2;
        }
        ++ans;
        int temp = tail;
        tail = 0;
        for(int i = 0;i < temp;++i){
            if(store[i].numb <= 0) continue;
            if(tail == 0 || store[i].color != store[tail - 1].color) store[tail++] = store[i];
            else store[tail - 1].numb += store[i].numb;
        }
    }
    printf("%d
",ans);
    return 0;
}

/*
zaaaabcdaaaaz
*/