问题 A: Super-palindrome
时间限制: 1 Sec 内存限制: 128 MB
提交: 560 解决: 210
[提交] [状态] [命题人:admin]
题目描述
You are given a string that is consisted of lowercase English alphabet. You are supposed to change it into a super-palindrome string in minimum steps. You can change one character in string to another letter per step.
A string is called a super-palindrome string if all its substrings with an odd length are palindrome strings. That is, for a string s, if its substring si...j satisfies j - i + 1 is odd then si+k = sj-k for k = 0,1,...,j-i+1.
输入
The fi rst line contains an integer T (1≤T≤100) representing the number of test cases.
For each test case, the only line contains a string, which consists of only lowercase letters. It is guaranteed that the length of string satisfies 1≤|s|≤100.
输出
For each test case, print one line with an integer refers to the minimum steps to take.
样例输入
复制样例数据
3 ncncn aaaaba aaaabb 样例输出 0 1 2 提示 For second test case aaaaba, just change letter b to a in one step.
题意：
给你一串小写字母组成的字符串。你需要用最少的步骤把这个字符串变成超级回文串。每个步骤你都可以把字符串里的一个字符变成其它字符。
超级回文串 ：一个字符串它的所有长度为奇数的子串都是回文串
思路：
就是要让奇数位上的都是同一个元素，偶数位上的都是同一元素，所以就可以，分别记录奇数位，和偶数位上出现的字母，分别找出奇数位出现最多的字母的数量max1,和偶数位出现最多的字母的数量max2,len-max1-max2就是要变换的字母的数量
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int vis1[30],vis2[30];
int main(){
int t;
scanf("%d",&t);
while(t--){
int max1=-1,max2=-1;
memset(vis1,0,sizeof(vis1));
memset(vis2,0,sizeof(vis2));
string s;
cin>>s;
int len=s.length() ;
for(int i=0;i<len;i++){
if(i%2==0)
vis2[s[i]-'a']++;
else
vis1[s[i]-'a']++;
}
for(int i=0;i<30;i++){
if(max1<vis1[i])
max1=vis1[i];
if(max2<vis2[i])
max2=vis2[i];
}
printf("%d\n",len-max1-max2);
}
return 0;
}
还可以动态的查找，需要变动的最小次数，每到一个字母，都要看在奇数（偶数）项里边，全部字母都变成这一种所需要的个数，每次都取最小，偶数最小count1,奇数最小count2，所求的=count1+count2
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int main(){
int t;
scanf("%d",&t);
while(t--){
string s;
cin>>s;
int len=s.length() ;
int count1=99999;
int count2=99999;
int count;
for(int i=0;i<len;i++){
count=0;
if(i%2==0){
for(int j=0;j<len;j+=2){
if(s[j]!=s[i])
count++;
}
count1=min(count1,count);
}
else{
for(int j=1;j<len;j+=2){
if(s[j]!=s[i])
count++;
}
count2=min(count2,count);
}
}
printf("%d\n",count1+count2);
}
return 0;
}