void printLCS(string str1, string str2, vector<vector<int> >flag, int idx1, int idx2)
{if(idx1 == 0 || idx2 == 0 ) return;if(flag[idx1][idx2] == 1){   printLCS(str1, str2,flag, idx1-1, idx2-1);cout << idx1 <<"\t"<< idx2 <<"\t";cout << str1[idx1-1] <<"\t"<<endl;}   else if(flag[idx1][idx2] == 2)printLCS(str1, str2,flag, idx1, idx2-1);else if(flag[idx1][idx2] == 3)printLCS(str1, str2,flag, idx1-1, idx2);
}int lcs(string str1, string str2)
{const size_t len1 = str1.size();const size_t len2 = str2.size();if(len1 == 0 || len2 == 0)return 0;int f[len1 + 1][len2 + 1];vector<vector<int> >flag;vector<int> tmp;tmp.resize(len2+1);for(size_t i = 0; i<= len1; i++)flag.push_back(tmp);//memset(flag,0,sizeof(flag)); // 1: leftup; 2: left; 3: upfor(size_t i = 0; i <= len1; i++){f[i][0] = 0;}for(size_t i = 0; i <= len2; i++){f[0][i] = 0;}for(size_t i = 1; i <= len1; i++){for(size_t j = 1; j <= len2; j++){if(str1[i-1] == str2[j-1]){f[i][j] = f[i-1][j-1] + 1;flag[i][j] = 1;}else{f[i][j] = max(f[i][j-1], f[i-1][j]);if(f[i][j-1] > f[i-1][j])flag[i][j] = 2;elseflag[i][j] = 3;}}}
#if 0for(size_t i = 1; i <= len1; i++){for(size_t j = 1; j <= len2; j++){//cout << "f["<<j<<"][" <<i<<"]" << f[j][i] <<"\n";cout << f[i][j] <<"\t";}cout << endl;}cout << endl;for(size_t i = 1; i <= len1; i++){for(size_t j = 1; j <= len2; j++){//cout << "f["<<j<<"][" <<i<<"]" << f[j][i] <<"\n";cout << flag[i][j] <<"\t";}cout << endl;}
#endifprintLCS(str1, str2, flag, len1, len2);return f[len1][len2];}

/**
找出两个字符串的最长公共连续子串的长度
** author :liuzhiwei
** data   :2011-08-16
**/
#include "stdio.h"
#include "string.h"
#include "stdlib.h"int longest_common_substring(char *str1, char *str2)
{int i,j,k,len1,len2,max,x,y;len1 = strlen(str1);len2 = strlen(str2);int **c = new int*[len1+1];for(i = 0; i < len1+1; i++)c[i] = new int[len2+1];for(i = 0; i < len1+1; i++)c[i][0]=0;        //第0列都初始化为0for(j = 0; j < len2+1; j++)c[0][j]=0;        //第0行都初始化为0 max = -1;for(i = 1 ; i < len1+1 ; i++){for(j = 1; j < len2+1; j++){if(str1[i-1]==str2[j-1])     //只需要跟左上方的c[i-1][j-1]比较就可以了c[i][j]=c[i-1][j-1]+1;else                         //不连续的时候还要跟左边的c[i][j-1]、上边的c[i-1][j]值比较，这里不需要c[i][j]=0;if(c[i][j]>max){max=c[i][j];x=i;y=j;}}}//输出公共子串char s[1000];k=max;i=x-1,j=y-1;s[k--]='\0';while(i>=0 && j>=0){if(str1[i]==str2[j]){s[k--]=str1[i];i--;j--;}else       //只要有一个不相等，就说明相等的公共字符断了，不连续了break;}printf("最长公共子串为：");puts(s);for(i = 0; i < len1+1; i++)         //释放动态申请的二维数组
        delete[] c[i];delete[] c;return max;
}
int main(void)
{char str1[1000],str2[1000];printf("请输入第一个字符串：");gets(str1);printf("请输入第二个字符串：");gets(str2);int len = longest_common_substring(str1, str2);printf("最长公共连续子串的长度为：%d\n",len);system("pause");return 0;
}