POJ2250 Compromise (LCS)

来源：互联网发布：linux怎么查看日志编辑：程序博客网时间：2024/04/28 00:20

要点：
1.比较基础的lcs题，关键在于输入时用字符串输入，输出时用递归

#include<stdio.h>#include<stdlib.h>#include<string.h>int len1, len2;char a[500][35], b[500][35];int num[500][500], flag[500][500];void lcs(){    int i, j;    for (i = 1; i <= len1; i++)    {        for (j = 1; j <= len2; j++)        {            if (strcmp(a[i-1], b[j-1]) == 0)            {                num[i][j] = num[i - 1][j - 1] + 1;                flag[i][j] = 1;            }            else if (num[i - 1][j]>num[i][j - 1])            {                num[i][j] = num[i - 1][j];                flag[i][j] = 2;  //标记向下            }            else            {                num[i][j] = num[i][j - 1];                flag[i][j] = 3; //标记向右            }        }    }}void getlcs(int i,int j)  //这里要用递归，用数组赋值会出现格式不匹配{    if (i == 0 || j == 0) //i或j触底        return;    if (flag[i][j] == 1)    {        getlcs(i - 1, j - 1);        printf("%s ", a[i - 1]); //注意这里是输出i-1    }    else if (flag[i][j] == 2)        getlcs(i - 1, j);    else        getlcs(i, j - 1);}int main(){    int i, j;    while (scanf("%s", a[0]) != EOF)  //先输入第一个    {  //用%s输入一个个字符串，比一个个字符输入减少时间        for (len1 = 1;; len1++)        {            scanf("%s", a[len1]);            if (strcmp(a[len1], "#") == 0)                break;        }        for (len2 = 0;; len2++)        {            scanf("%s", b[len2]);            if (strcmp(b[len2], "#") == 0)                break;        }        lcs();        getlcs(len1,len2);        printf("\n");    }    return 0;}

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