【KMP求多个串的最长公共子串】POJ 3450

最新推荐文章于 2022-11-22 23:25:33 发布

原创最新推荐文章于 2022-11-22 23:25:33 发布 · 800 阅读

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本文介绍了一道基于POJ3080修改的ACM竞赛题目的解决方案，采用KMP算法进行字符串匹配，通过动态规划求解最长公共前缀，并针对多个目标字符串进行匹配验证。

奇怪，这题是POJ3080修改而ac的，在POJ3080却过不了，很诡异的说。。。难道哥的RP低成这样？！

#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <time.h>
#include <cstdio>
#include <math.h>
#include <iomanip>
#include <cstdlib>
#include <limits.h>
#include <string.h>
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

#define LL long long
#define MIN INT_MIN
#define MAX INT_MAX
#define PI acos(-1.0)
#define FRE freopen("input.txt","r",stdin)
#define FF freopen("output.txt","w",stdout)
#define N 205
char str[4005][N];
int p[N];
int m;
int l[N];
bool chk(char *ss) {
    int i,j;
    int len = strlen(ss+1);
    p[1] = 0;
    j = 0;
    for (i = 2; i <= len; i++) {
        while (j > 0 && ss[i] != ss[j+1]) {
            j = p[j];
        }
        if (ss[i] == ss[j+1]) j++;
        p[i] = j;
    }
    for (int kk = 2; kk <= m; kk++) {
        j = 0;
        for (i = 1; i <= l[kk]; i++) {
            while (j > 0 && ss[j+1] != str[kk][i]) {
                j = p[j];
            }
            if (ss[j+1] == str[kk][i]) j++;
            if (j == len) break;
        }
        if (i > l[kk])return false;
    }
    return true;
}

int main () {
    while (scanf("%d",&m) && m) {
        int i,j;//
        for (i = 1; i <= m; i++) {
            scanf("%s",str[i]+1);
            l[i] = strlen(str[i]+1);
            //printf("!%s\n",str[i]+1);
        }
        char ss[N],ans[N];
        int maxlen = 0;
        memset(ans,'\0',sizeof(ans));
        for (int len = 1; len <= l[1]; len++) {
            for (i = 1; i + len <= l[1] + 1; i++) {
                int cnt = 1;
                for (j = i; j  <= len + i - 1; j++) {
                    ss[cnt++] = str[1][j];
                }
                ss[cnt] = '\0';//printf("!%s\n",ss+1);
                if (chk(ss)) {
                        if (len == maxlen) {
                            if (strcmp(ans+1,ss+1) > 0) {
                                strcpy(ans+1,ss+1);
                            }
                        } else if (len > maxlen) {
                            maxlen = len;
                            strcpy(ans+1,ss+1);
                        }
                }
            }
        }
        if (strlen(ans+1) == 0) puts("IDENTITY LOST");
        else printf("%s\n",ans+1);
    }
    return 0;
}