字符串

来源：互联网发布：网络喷子都是穷人编辑：程序博客网时间：2024/06/08 03:11

近似回文词

Problem's Link：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1328

analyse:

直接暴力枚举每一个终点，然后枚举回文串的半径即可.

Time complexity：O(n*m)

Source code：

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-10-03-14.25#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 1000010#define LL long longusing namespace std;int k,real[1100],sta,max_len,cas=1;char st[1100],ss[1100];int main(){    ios_base::sync_with_stdio(false);    cin.tie(0);//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);    while(~scanf("%d",&k))    {        getchar();        gets(st);        int len=0;        max_len=0;        int l1=strlen(st);        for(int i=0;i<l1;++i)        {            if((st[i]>='a'&&st[i]<='z')||(st[i]>='A'&&st[i]<='Z'))            {                if(st[i]>='A'&&st[i]<='Z')                    st[i]+=32;                ss[len]=st[i];                real[len]=i;                len++;            }        }        for(int i=0;i<len;++i)        {            int error=0,j;            for(j=0;i+j<len&&i-j>=0;++j)            {                if(ss[i+j]!=ss[i-j])                    error++;                if(error>k)                    break;            }            j--;            if(real[i+j]-real[i-j]+1>max_len)            {                max_len=real[i+j]-real[i-j]+1;                sta=real[i-j];            }            error=0;            for(j=1;i+j<len&&i-j+1>=0;++j)            {                if(ss[i+j]!=ss[i-j+1])                    error++;                if(error>k)                    break;            }            j--;            if(j<=0) continue;            if(real[i+j]-real[i-j+1]+1>max_len)            {                max_len=real[i+j]-real[i-j+1]+1;                sta=real[i-j+1];            }        }        printf("Case %d: %d %d\n",cas++,max_len,sta+1);    }    return 0;}

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