HDU 5455 Fang Fang（模拟）

最新推荐文章于 2018-12-01 19:48:22 发布

原创最新推荐文章于 2018-12-01 19:48:22 发布 · 651 阅读

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本文介绍了一个字符串匹配的问题，通过遍历和条件判断实现解决方案。输入为一系列字符串，目标是最小化使用特定子串拼接原串所需的子串数量，或判断是否无法完成拼接。

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思路：直接模拟就好了，注意全是f的情况

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 1000000 + 100;
char s[maxn];
int pos[maxn];

int main()
{
	int T, cas=1, n;
	scanf("%d", &T);
	while(T--)
	{
		int f=1, cn=0;
		scanf("%s", s);
		n = strlen(s);
		for(int i=0;s[i];++i)
		{
			if(s[i]=='c') pos[cn++]=i;
			if(s[i]!='c'&&s[i]!='f'){
				f=0; break;
			}
		}
		printf("Case #%d: ", cas++);

		if(!f) puts("-1");
		else
		{
			if(cn==0)
			{
				printf("%d\n", n/2+n%2);
				continue;
			}

			for(int i=0;i<cn-1;++i)
			{
				if(pos[i+1]-pos[i]<3)
				{
					f=0; break;
				}
			}
			if(n-pos[cn-1] + pos[0] < 3) f=0;
			if(!f) puts("-1");
			else printf("%d\n", cn);
		}

	}
	
	return 0;
}

Description

Fang Fang says she wants to be remembered.
I promise her. We define the sequence

of strings.

Write down a serenade as a lowercase string

in a circle, in a loop that never ends.
Spell the serenade using the minimum number of strings in

, or nothing could be done but put her away in cold wilderness.

Input

An positive integer

, indicating there are

test cases.
Following are

lines, each line contains an string

as introduced above.
The total length of strings for all test cases would not be larger than

Output

The output contains exactly

lines.
For each test case, if one can not spell the serenade by using the strings in

, output

. Otherwise, output the minimum number of strings in

to split

according to aforementioned rules. Repetitive strings should be counted repeatedly.

Sample Input

8
ffcfffcffcff
cffcfff
cffcff
cffcf
ffffcffcfff
cffcfffcffffcfffff
cff
cffc

Sample Output

Case #1: 3
Case #2: 2
Case #3: 2
Case #4: -1
Case #5: 2
Case #6: 4
Case #7: 1
Case #8: -1