Usaco1.1 beads

最新推荐文章于 2024-11-18 20:32:13 发布

cnyali_ljf

最新推荐文章于 2024-11-18 20:32:13 发布

阅读量573

点赞数 1

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/cnyali_ljf/article/details/54835759

本文介绍了一种解决珠子串匹配问题的算法实现，通过两次扫描寻找最长匹配串，适用于黑白两色珠子的配对问题。算法首先从左侧开始扫描，记录最长连续匹配的珠子数，再从右侧重复这一过程，最终确定最长匹配长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
ID: ljfcnya1
PROG: beads
LANG: C++
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>

using namespace std;

#define REP(i, a, b) for(int i = (a), _end_ = (b);i <= _end_; ++ i)
#define mem(a) memset((a), 0, sizeof(a))
#define str(a) strlen(a)

const int maxn = 10000;

int n, m;

int ans[maxn];

string s;

int main() {
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    scanf("%d", &n);
    getchar();
    cin>>s;
    s += s;
    REP(i, 0, n) {
        int j = i - 1;
        while(s[++j] == 'w');
        char tmp = s[j];
        while((s[++j] == tmp || s[j] == 'w') && (j >= i && j < i + n));
        ans[i] += j - i;
        if(ans[i] == n)
            continue;   
        j = i + n;
        while(s[--j] == 'w');
        tmp = s[j];
        while((s[--j] == tmp || s[j] == 'w') && (j >= i && j < i + n));
        ans[i] += i + n - j - 1;
        if(ans[i] > n)  ans[i] = n;
    }
    int sum = 0;
    REP(i, 1, n)
        sum = max(sum, ans[i]);
    printf("%d\n", sum);
    return 0;
}