USACO 4.3 Street Race(DFS)

本来想用floyd的，利用p[i][j] < INF&&p[j][i] < INF 来判断的，提交发现在前边或者后边的跑道存在环的时候，这样做是错的。。。我改了一下，直接暴力。。。

  1 /*

  2 ID:cuizhe

  3 LANG: C++

  4 TASK: race3

  5 */

  6 #include <cstdio>

  7 #include <cmath>

  8 #include <cstring>

  9 #include <iostream>

 10 #include <map>

 11 #include <queue>

 12 #include <algorithm>

 13 using namespace std;

 14 #define INF 100000

 15 int p[101][101];

 16 int o[101];

 17 int flag[101];

 18 int n;

 19 int dfs(int x)

 20 {

 21     int i;

 22     if(x == n) return 1;

 23     for(i = 0;i <= n;i ++)

 24     {

 25         if(p[x][i] == 1&&!o[i])

 26         {

 27             o[i] = 1;

 28             if(dfs(i))

 29             return 1;

 30             o[i] = 0;

 31         }

 32     }

 33     return 0;

 34 }

 35 void dfs1(int x,int key)

 36 {

 37     int i;

 38     if(x == key) return;

 39     for(i = 0;i <= n;i ++)

 40     {

 41         if(p[x][i] == 1&&!o[i])

 42         {

 43             o[i] = 1;

 44             dfs1(i,key);

 45         }

 46     }

 47     return ;

 48 }

 49 int dfs2(int x,int key)

 50 {

 51     int i;

 52     if(x == key) return 0;

 53     for(i = 0;i <= n;i ++)

 54     {

 55         if(p[x][i] == 1&&!o[i])

 56         {

 57             o[i] = 2;

 58             if(dfs2(i,key))

 59             return 1;

 60         }

 61         else if(p[x][i] == 1&&o[i] == 1)

 62         {

 63             return 1;

 64         }

 65     }

 66     return 0;

 67 }

 68 int main()

 69 {

 70     int i,j,num,ans = 0;

 71     freopen("race3.in","r",stdin);

 72     freopen("race3.out","w",stdout);

 73     for(i = 0;i <= 50;i ++)

 74     {

 75         for(j = 0;j <= 50;j ++)

 76         p[i][j] = INF;

 77         p[i][i] = 0;

 78     }

 79     for(i = 0;;i ++)

 80     {

 81         scanf("%d",&num);

 82         if(num == -1) break;

 83         if(num == -2) continue;

 84         p[i][num] = 1;

 85         for(;;)

 86         {

 87             scanf("%d",&num);

 88             if(num == -2) break;

 89             p[i][num] = 1;

 90         }

 91     }

 92     n = i-1;

 93     for(i = 1;i < n;i ++)

 94     {

 95         memset(o,0,sizeof(o));

 96         o[i] = 1;

 97         if(!dfs(0))

 98         {

 99             flag[i] = 1;

100             ans ++;

101         }

102         o[i] = 0;

103     }

104     printf("%d",ans);

105     for(i = 1;i < n;i ++)

106     {

107         if(flag[i]) printf(" %d",i);

108     }

109     printf("\n");

110     ans = 0;

111     for(i = 1;i < n;i ++)

112     {

113         if(flag[i])

114         {

115             memset(o,0,sizeof(o));

116             o[0] = 1;

117             dfs1(0,i);

118             o[i] = 2;

119             if(dfs2(i,n))

120             continue;

121             else

122             {

123                 flag[i] = 2;

124                 ans ++;

125             }

126         }

127     }

128     printf("%d",ans);

129     for(i = 1;i < n;i ++)

130     {

131         if(flag[i] == 2) printf(" %d",i);

132     }

133     printf("\n");

134     return 0;

135 }