kmp poj 3450 3461

这两题都比较简单，很明显的kmp求解

poj 3450

题目大意：就是求最长的公共子串，如果有长度一样的，取字典序最小的

跟poj 3080基本一模一样，没啥好说的~

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 4010;const int maxm = 210;int n, next[maxm];char str[maxn][maxm];void get_next(int start, int len);bool compare(int start, int len, int index);int main(){        while(true)    {        scanf("%d", &n);        if(n == 0)            break;        bool inital = true;        char ans[maxm] = "IDENTITY LOST";        for(int i = 0; i < n; i++)            scanf("%s", str[i]);                    int len = strlen(str[0]);        bool sign = false;        for(int i = len; i >= 1; i--)        {            for(int j = 0; j + i <= len; j++)            {                get_next(j, i);                bool flag = true;                for(int k = 1; k < n; k++)                {                    if(!compare(j, i, k))                    {                        flag = false;                        break;                    }                }                if(flag)                {                    if(inital || strncmp(ans, str[0]+j, i) > 0)                    {                        inital = false;                        strncpy(ans, str[0]+j, i);                        ans[i] = '\0';                    }                    sign = true;                }            }            if(sign)                break;        }        printf("%s\n", ans);    }        return 0;}void get_next(int start, int len){    int pos = start - 1;    int i = start, j = pos, end = start + len;    next[start] = pos;    while(i < end)    {        if(j == pos || str[0][i] == str[0][j])        {            i++; j++;            next[i] = j;        }        else            j = next[j];    }}bool compare(int start, int len, int index){    int pos = start - 1;    int i = start, j = 0, end = start + len;    int length = strlen(str[index]);    while(i < end && j < length)    {        if(i == pos || str[0][i] == str[index][j])        {            i++; j++;        }        else            i = next[i];    }    if(i == end)        return true;    return false;}

poj 3461

这题也很简单，水题~

题目大意：给几组字符串数据，要你求匹配串在模式串里面出现了几次

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 1000010;const int maxm = 10010;int next[maxm], n;char str[maxn], p[maxm];int main(){    scanf("%d", &n);    int i, j, plen, slen, ans;    while(n-- != 0)    {        ans = 0;        scanf("%s", p);        scanf("%s", str);        plen = strlen(p);        slen = strlen(str);        i = 0, j = -1;        next[0] = -1;        while(i < plen)        {            if(j == -1 || p[i] == p[j])            {                i++; j++;                next[i] = j;            }            else                j = next[j];        }        i = 0; j = 0;        while(j < slen)        {            if(i == -1 || p[i] == str[j])            {                i++; j++;            }            else                i = next[i];                            if(i == plen)            {                ans++;                i = next[i];            }        }        printf("%d\n", ans);    }        return 0;}