uva 140 注释超多

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uDebug

Time limit

3000 ms

Linux

Result	Time
Accepted	3 min ago
Time limit exceeded	6 min ago
Time limit exceeded	11 min ago
Accepted	13 min ago
Compilation error	17 min ago
Time limit exceeded	18 min ago
Time limit exceeded	20 min ago
Accepted	2 hr ago
Wrong answer	2 hr ago
Wrong answer	2 hr ago
More...

Descriptions:

System Crawler 2018-07-04

Bandwidth

UVA - 140

Preview:

Post & BroadcastPostClear

Status	Accepted
Time	30ms
Length	1639
Lang	C++11 5.3.0
Submitted	2018-07-08 14:19:17
Shared
RemoteRunId	21564153

Select Code

#include<stdio.h>
#include<string.h>
#include<ctype.h>
using namespace std;
int  n,max,map[26][26],visit[26];
char s[10],s1[10];  //用字符串保存结果待会比较字典序用库函数比较方便
int dfs(int m)
{
    int i,j,min;
    if (m>=n)
    {
        min=0;
        for (i=0; i<n-1; i++)
            for (j=i+1; j<n; j++)
                if ((map[s1[i]-'A'][s1[j]-'A']==1)&&(j-i>min))
                    min=j-i;//
        if (min<max)
        {
            max=min;
            strcpy(s,s1);
        }
        if ((min==max)&&(strcmp(s,s1)>0))//S>s1;
            strcpy(s,s1);//将s1赋值给s；
        return 0;
    }
    for (i=0; i<26; i++)//将所有结点排序，且是字典顺序。（重点）
        if (visit[i]==1)
        {
            visit[i]=0;
            s1[m]=i+'A';
            dfs(m+1);
            visit[i]=1;
        }
}
int main()
{
    char ch,CH;
    int i,j;
    while (scanf("%c",&ch)&&ch!='#')
    {
        memset(map,0,sizeof(map));
        memset(visit,0,sizeof(visit));
        //用二维数组来存放边，坐标为点
        while (scanf("%c",&CH)&&CH!='\n')
        {
            if (isalpha(CH))//是否为字母
            {
                map[ch-'A'][CH-'A']=1;
                map[CH-'A'][ch-'A']=1;
                visit[ch-'A']=1;
                visit[CH-'A']=1;
            }
            if (CH==';')
                scanf("%c",&ch);//很巧妙
        }
        n=0;//结点个数

        max=9999;
        for (i=0; i<26; i++)
            if (visit[i])//看看哪些点存在
            {
                s[n]=i+'A';
                ++n;
            }
        s[n]='\0';
        s1[n]='\0';
        dfs(0);
        for (i=0; s[i]!='\0'; i++)
            printf("%c ",s[i]);
        printf("-> %d\n",max);
    }
    return 0;
}