Fork me on GitHub

Uva 10596 - Morning Walk

 

Problem H

Morning Walk

Time Limit

3 Seconds

 

 

 

 

 

 

 

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1537

Problemsetter: Muhammad Abul Hasan

International Islamic University Chittagong

 解题思路：

感觉跟上一题一样，都是欧拉回路，这次不用打印较为简单，感觉欧拉回路开始有个模板了，判断结点度数的奇偶，DFS或者BFS或者并查集判断连通。

刚开始理解错题目了，以为可以是欧拉回路或通路，而且看他的输入以为是有向图，差的太远了，故WA了几次~~但改了之后还是WA了很多次，逃了一节课找了很久还是没找出来，最后干脆重新写过 AC了，感觉弱爆了！

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAXN 220
 4 int road[MAXN][MAXN];
 5 int visit[MAXN];
 6 int note[MAXN];
 7 int sum;
 8 
 9 int DFS(int current, int n)
10 {
11     int j;
12     sum++;
13     visit[current] = 1;
14     for(j=0; j<n; ++j)
15     {
16         if(road[current][j]  && !visit[j])
17         DFS(j, n);
18     }
19     return 1;
20 }
21 
22 int main()
23 {
24     int i, j, n, r, flag, x, y;
25     while(scanf("%d%d", &n, &r) != EOF)
26     {
27         memset(road, 0, sizeof(road));
28         memset(note, 0, sizeof(note));
29         for(i=0; i<r; ++i)
30         {
31             scanf("%d%d", &x, &y);
32             road[x][y]++;
33             road[y][x]++;
34             note[x]++, note[y]++;
35         }
36         flag = 0;
37         for(i=0; i<n; ++i)
38         {
39             if(note[i]%2)
40             {
41                 flag = 1;
42                 break;
43             }
44         }
45         if(flag)
46         {
47             printf("Not Possible\n");
48         }
49         else
50         {
51             memset(visit, 0, sizeof(visit));
52             sum = 0;
53             DFS(0, n);
54             if(sum != n)
55             printf("Not Possible\n");
56             else printf("Possible\n");
57         }    
58     }
59     return 0;
60 }
WA的代码 
#include<stdio.h>
#include<string.h>
#define MAXN 202
int road[MAXN][MAXN];
int note[MAXN][2];
int visit[MAXN];

int Dfs(int current, int n)
{
    int j;
    visit[current] = 1;
    for(j=0; j<n; ++j)
    if(note[current][j] != 0 && !visit[j])
    Dfs(j, n);
}
int main()
{
    freopen("input.txt", "r", stdin);
    int n, r, i, j, x, y, flag, cnt, temp,store;
    while(scanf("%d%d", &n, &r) != EOF)
    {
        memset(road, 0, sizeof(road));
        memset(note, 0, sizeof(note));
        for(i=0; i<r; ++i)
        {
            scanf("%d%d", &x, &y);
            road[x][y]++;
            note[x][0]++, note[y][1]++;
        }
        cnt = flag = store = 0;
        for(i=0; i<n; ++i)
        if(note[i][0] != 0)
        {
            memset(visit, 0, sizeof(visit));
            Dfs(i, n);
             for(j=0; j<n; ++j)
             if(!visit[j] && (note[j][0] != 0 || note[j][1] != 0))
             flag = 1;
            break;
        }
        for(i=0; i<n && flag == 0; ++i)
        {
            if(note[i][0] != 0 || note[i][1] != 0)
            {
                temp = note[i][0] - note[i][1];
                if(temp == -1 || temp == 1)
                {
                    cnt++;
                    store += temp;
                }
                else if(temp != 0) 
                {
                    flag = 1;
                    break;
                }
            }
            if(flag == 1) break;
        }
        if(flag == 1 || cnt > 2 || store != 0)
        {
            printf("Not Possible\n");
        }
        else printf("Possible\n");
    }
    return 0;
}

 

posted @ 2013-04-11 16:08  Gifur  阅读(347)  评论(0)  编辑  收藏  举报
TOP