hdu 1462 Word Crosses

題目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1462

題目描述：

Word Crosses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 287    Accepted Submission(s): 93

Problem Description

A `word cross' is formed by printing a pair of words, the first horizontally and the second vertically, so that they share a common letter. A `leading word cross' is one where the common letter is as near as possible to the beginning of the horizontal word, and, for this letter, as close as possible to the beginning of the vertical word. Thus DEFER and PREFECT would cross on the first 'E' in each word, PREFECT and DEFER would cross on the 'R'. `Double leading word crosses' use two pairs of words arranged so that the two horizontal words are on the same line and each pair forms a leading word cross.

Write a program that will read in sets of four words and form them (if possible) into double leading word crosses.

 

Input

Input will consist of a series of lines, each line containing four words (two pairs). A word consists of 1 to 10 upper case letters, and will be separated from its neighbours by at least one space. The file will be terminated by a line consisting of a single #.

 

Output

Output will consist of a series of double leading word crosses as defined above. Leave exactly three spaces between the horizontal words. If it is not possible to form both crosses, write the message `Unable to make two crosses'. Leave 1 blank line between output sets.

 

Sample Input

MATCHES CHEESECAKE PICNIC EXCUSES PEANUT BANANA VACUUM GREEDY A VANISHING LETTER TRICK #

 

Sample Output

C H E E S E E C X MATCHES PICNIC K U E S E S Unable to make two crosses V A LETTER N R I I S C H K I N G

題意：

分別讓兩個串交叉輸出。

題解：

此題特別注意格式，可以算出各自的長度，公共字符的位置，開始輸出串的位置等等都可以根據數值算出來。另外此題除了容易WA外也很容易PE，注意不要輸出多餘的空格，否則很容易PE

代碼：

#include<stdio.h>
#include<string.h>
char s1[12],s2[12],s3[12],s4[12];
int l1,l2,l3,l4;
int f1,f2,f3,f4;//the first letter 's position in the global coordinate
int c1,c2,c3,c4;//the common letter 's position in own string
int hlen=3, vlen, ilen;//number of spaces, initialize length of spaces for first column
int print_space(int num)
{
    for(int i=0;i<=num-1;i++) printf(" ");
    return(0);
}
int main()
{
    int cases = 0;
    while(scanf("%s",s1)!=EOF&&s1[0]!='#')
    {
        if(cases > 0) printf("\n");
        int i,j;
        int line=0;//the length of coordinate
        scanf("%s%s%s",s2,s3,s4);
        l1=strlen(s1);
        l2=strlen(s2);
        l3=strlen(s3);
        l4=strlen(s4);
        int flag1=0;//can not print result
        for(i=0;i<=l1-1;i++)
        {
            for(j=0;j<=l2-1;j++)
            {
                if(s1[i] == s2[j])
                {
                    c1=i;
                    c2=j;
                    flag1=1;
                    break;
                }
            }
            if(j<=l2-1) 
            {
                break;
            }
        }
        int flag2=0;
        for(i=0;i<=l3-1;i++)
        {
            for(j=0;j<=l4-1;j++)
            {
                if(s3[i] == s4[j])
                {
                    c3=i;
                    c4=j;
                    flag2=1;
                    break;
                }
            }
            if(j<=l4-1) 
            {
                break;
            }
        }
        if(flag1==1&&flag2==1)
        {
            if(c2 > c4)//f2=0
            {
                f1=c2;
                f2=0;
                f3=f1;
                f4=f3-c4;   
                line = f4+l4-1 > l2-1 ? f4+l4-1 : l2-1;
            }
            else
            {
                f4=0;
                f1=c4;
                f3=f1;
                f2=f3-c2;
                line = f2+l2-1 > l4-1 ? f2+l2-1 : l4-1;
            }
            vlen=l1-1-c1 + 3 + c3;
            ilen=c1;
            for(i=0;i<=line;i++)
            {
                if(i!=f1)
                {
                    print_space(ilen);
                    if(i>=f2&&(i-f2)<=l2-1) printf("%c",s2[i-f2]);
                    else printf(" ");
                }
                if(i!=f1)
                {
                    if(i>=f4&&(i-f4)<=l4-1) print_space(vlen);//PE
                    if(i>=f4&&(i-f4)<=l4-1) printf("%c",s4[i-f4]);
                    //else printf(" ");//PE
                }
                if(i==f1)
                {
                    printf("%s",s1);
                    print_space(hlen);
                    printf("%s",s3);
                }
                printf("\n");
            }
        }
        else//print no answer
        {
            printf("Unable to make two crosses\n");
        }
        cases++;
    }
    return(0);
}