POJ2503：Babelfish(二分)

标签：poj   二分   

Description

Input

Output

Sample Input

dog ogday
cat atcay
pig igpay
froot ootfray
loops oopslay

atcay
ittenkay
oopslay

Sample Output

cat
eh
loops

咋看是字典树，但是我在搜二分专题的时候看到的这道题，那么这道题肯定能用二分解决

思路很好想，运用到了sscanf函数

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

struct node
{
    char s1[20],s2[20];
} a[100005];
int len;
int cmp(node a,node b)
{
    return strcmp(a.s2,b.s2)<0;
}

int main()
{
    len = 0;
    int i,j;
    char str[50];
    while(gets(str))
    {
        if(str[0] == '\0')
        break;
        sscanf(str,"%s%s",a[len].s1,a[len].s2);
        len++;
    }
    sort(a,a+len,cmp);
    while(gets(str))
    {
        int l = 0,r= len-1,mid,flag = 1;
        while(l<=r)
        {
            int mid = (l+r)>>1;
            if(strcmp(str,a[mid].s2)==0)
            {
                printf("%s\n",a[mid].s1);
                flag = 0;
                break;
            }
            else if(strcmp(str,a[mid].s2)<0)
                r = mid-1;
            else
                l = mid+1;
        }
        if(flag)
            printf("eh\n");
    }

    return 0;
}

POJ2503：Babelfish(二分),布布扣,bubuko.com

POJ2503：Babelfish(二分)

标签：poj   二分   

原文地址：http://blog.csdn.net/libin56842/article/details/38307085