wenbao与manacher

推荐博客：https://segmentfault.com/a/1190000003914228

 

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模板

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn = 1e6+10;
char str[2*maxn];
int b[2*maxn];

int main() {
    int n;
    scanf("%d", &n);
    while(n--) {
        scanf("%s", str);
        int len = strlen(str), j = 0, ma = 0;
        for(int i = len; i >= 0; i--) {
            str[2*i+2] = str[i];
            str[i*2+1] = '#';
        }
        str[0] = '@';
        for(int i = 2; i <2*len+1; i++) {
            if(j+b[j] > i) b[i] = min(b[2*j-i], j+b[j]-i); //i在pos左边
            else b[i] = 1;//i在pos右边
            while(str[i-b[i]] == str[i+b[i]]) b[i]++;
            if(j+b[j] < i+b[i]) j = i;//最长的中心位置
            if(ma < b[i]) ma = b[i];//回文串的最长长度+1
        }
        printf("%d
", ma-1);
    }
    return 0;
}

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小刘

#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 1e6+10;
char str[maxn];
int main(){
    int n;
    cin>>n;
    while( n -- ){
    memset( str, 0, sizeof( str ) );
    cin>>str;
    int len = strlen(str),man = -1,j;
    for( int i = 0; i < len; i ++ ){
        for( j = 0; i - j >= 0 && i + j < len; j ++ ){
            if( str[i-j] != str[i+j] ) break;
        }
        man = max( man, 2 * j - 1) ;
        for( j = 0; i - j >= 0 && i + j + 1 < len; j ++ ){
            if( str[i - j] != str[i + j + 1] ) break;
        }
        man = max( man, 2 * j  );
    }
    cout<<man<<endl;
    }
}

输出最大回文子串

#include <iostream>
#include <ctype.h>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn = 110000+10;
char str[maxn], s[maxn];
int p[maxn];
int main(){
    int len, m = 0, man = 0, x, y;
    int i, j;
    fgets(str, 1000, stdin); //fgets(s, n, stdin); 从键盘读入长度为不大于n的字符完美替代gets
    // cout<<s<<endl;
    len = strlen(str);
    for(i = 0; i < len; i ++){
        if(isalpha(str[i])){
            p[m] = i;
            s[m++] = toupper(str[i]);
        }
    }
    // cout<<s<<endl;
    // cout<<m<<endl;
    for(i = 0; i < m; i ++){
        for(j = 0; i - j >= 0 && i + j < m; j ++){
            if(s[i-j] != s[i+j]) break;
            if(man < 2*j+1) man = 2*j+1, x = p[i-j], y = p[i+j];
        }
        for(j = 0; i - j >= 0 && i + j + 1 < m; j ++){
            if(s[i - j] != s[i + j + 1]) break;
        if(man < 2*j+2) man = 2*j+2, x = p[i-j], y = p[i+j+1]; 
        }
    }
    for(i = x; i <= y; i ++){
        printf("%c", str[i]);
    }
    printf("
");
    return 0;
}

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http://acm.timus.ru/problem.aspx?space=1&num=1297

 输出最长回文串

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

const int maxn=1e3+10;
char str[2*maxn], ss[maxn];
int b[2*maxn];

int main(){
    int n;
    scanf("%s",ss);
    int len = strlen(ss), j = 0, ma = 0, id;
    for(int i = len; i >= 0; i--){
        str[2*i+2]=ss[i];
        str[i*2+1]='#';
    }
    str[0]='@';
    for(int i = 2; i < 2*len+1; i++){
        if(j+b[j] > i) b[i] = min(b[2*j-i], j+b[j]-i);
        else b[i] = 1;
        while(str[i-b[i]] == str[i+b[i]]) b[i]++;
        if(j+b[j] < i+b[i]) j = i;
        if(ma < b[i]) ma = b[i], id = j/2;
    }
    //printf("%d %d
", id, ma);
    ma--;
    id = id - ma/2 + (ma%2 == 0);
    ss[id+ma-1] = 0;
    //printf("%d
", id);
    //printf("%d
",ma);
    printf("%s
", ss+id-1);
    return 0;   
}

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只有不断学习才能进步！