朗讯杯

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2718

模拟 

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 using namespace std;

  6 int w[25][25];

  7 int dis[4][2] = {1,0,0,-1,0,1,-1,0};

  8 int main()

  9 {

 10     int n,m,k,i,s1,s2,kk=0,d;

 11     while(cin>>n>>m>>k)

 12     {

 13         memset(w,0,sizeof(w));

 14         kk++;

 15         if(!n&&!m&&!k) break;

 16         for(i = 1; i <= k ; i++)

 17         {

 18             int x,y;

 19             cin>>x>>y;

 20             w[x][y] = 1;

 21         }

 22         cin>>s1>>s2;

 23         int num=0;

 24         if(s1==1)

 25         d = 0;

 26         else if(s2==1)

 27         d = 2;

 28         else if(s1==n)

 29         d = 3;

 30         else if(s2==m)

 31         d = 1;

 32         printf("Case %d: ",kk);

 33         int tx = s1,ty=s2,dd=d;

 34         num=1;

 35         while(1)

 36         {

 37             if(tx==1||ty==1||tx==n||ty==m)

 38             {

 39                 if(tx==1&&d==3)

 40                 {printf("%d %d %d\n",tx,ty,num);break;}

 41                 else if(ty==1&&d==1)

 42                 {printf("%d %d %d\n",tx,ty,num);break;}

 43                 else if(tx==n&&d==0)

 44                 {printf("%d %d %d\n",tx,ty,num);break;}

 45                 else if(ty==m&&d==2)

 46                 {printf("%d %d %d\n",tx,ty,num);break;}

 47             }

 48             int ox = dis[d][0]+tx;

 49             int oy = dis[d][1]+ty;

 50             if(w[ox][oy])

 51             {

 52                 if(d==0)

 53                 {

 54                     d = 1;

 55                     ox = dis[d][0]+tx;

 56                     oy = dis[d][1]+ty;

 57                     if(w[ox][oy])

 58                     {

 59                         d = 2;

 60                         ox = dis[d][0]+tx;

 61                         oy = dis[d][1]+ty;

 62                         if(w[ox][oy])

 63                         d = 3;

 64                     }

 65                 }

 66                 else if(d==1)

 67                 {

 68                     d = 3;

 69                     ox = dis[d][0]+tx;

 70                     oy = dis[d][1]+ty;

 71                     if(w[ox][oy])

 72                     {

 73                         d = 0;

 74                         ox = dis[d][0]+tx;

 75                         oy = dis[d][1]+ty;

 76                         if(w[ox][oy])

 77                         d = 2;

 78                     }

 79                 }

 80                 else if(d==3)

 81                 {

 82                     d = 2;

 83                     ox = dis[d][0]+tx;

 84                     oy = dis[d][1]+ty;

 85                     if(w[ox][oy])

 86                     {

 87                         d = 1;

 88                         ox = dis[d][0]+tx;

 89                         oy = dis[d][1]+ty;

 90                         if(w[ox][oy])

 91                         d = 0;

 92                     }

 93                 }

 94                 else

 95                 {

 96                     if(d==2)

 97                     {

 98                         d = 0;

 99                         ox = dis[d][0]+tx;

100                         oy = dis[d][1]+ty;

101                         if(w[ox][oy])

102                         {

103                             d = 3;

104                             ox = dis[d][0]+tx;

105                             oy = dis[d][1]+ty;

106                             if(w[ox][oy])

107                             d = 1;

108                         }

109                     }

110                 }

111             }

112             else

113             {

114                 num++;

115                 tx = ox;

116                 ty = oy;

117             }

118         }

119     }

120     return 0;

121 }

122  

123 

124 

125 

126 /**************************************

127     Problem id    : SDUT OJ 2718 

128     User name    : shang 

129     Result        : Accepted 

130     Take Memory    : 472K 

131     Take Time    : 0MS 

132     Submit Time    : 2013-12-01 10:48:41  

133 **************************************/

View Code

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2719

水题

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<queue>

 7 #include<vector>

 8 using namespace std;

 9 int ma[3][1010];

10 int main()

11 {

12     int n,m,k,a,kk=0,i;

13     while(cin>>n>>m>>k)

14     {

15         if(!n&&!m&&!k)

16         break;

17         memset(ma,0,sizeof(ma));

18         kk++;

19         for(i = 1; i <= k ; i++)

20         {

21             scanf("%d",&a);

22             int k = a%n,t = 1,q = a;

23             while(ma[t][k])

24             {

25                 int y = ma[t][k];

26                 ma[t][k] = q;

27                 q = y;

28                 if(t==1)

29                 {

30                     t = 2;

31                     k = y%m;

32                 }

33                 else

34                 {

35                     t = 1;

36                     k = y%n;

37                 }

38             }

39             ma[t][k] = q;

40         }

41         printf("Case %d:\n",kk);

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

43         {

44             if(ma[1][i])

45             break;

46         }

47         if(i!=n)

48         puts("Table 1");

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

50         {

51             if(ma[1][i])

52             printf("%d:%d\n",i,ma[1][i]);

53         }

54         for(i = 0 ; i < m ; i++)

55         {

56             if(ma[2][i])

57             break;

58         }

59         if(i!=m)

60         puts("Table 2");

61         for(i = 0 ; i < m ; i++)

62         {

63             if(ma[2][i])

64             printf("%d:%d\n",i,ma[2][i]);

65         }

66     }

67     return 0;

68 }

69  

70 

71 

72 

73 /**************************************

74     Problem id    : SDUT OJ 2719 

75     User name    : shang 

76     Result        : Accepted 

77     Take Memory    : 500K 

78     Take Time    : 0MS 

79     Submit Time    : 2013-12-01 11:12:16  

80 **************************************/

View Code

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2721

模拟 消行消列 bfs去元素

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<queue>

  7 #include<vector>

  8 

  9 using namespace std;

 10 int a[15][15],vis[15][15];

 11 int dis[4][2] = {0,1,0,-1,1,0,-1,0};

 12 struct node

 13 {

 14     int x,y;

 15 };

 16 void init()

 17 {

 18     int i,j;

 19     for(i = 2; i <= 12 ; i++)

 20     {

 21         for(j = 1; j <= 10 ; j++)

 22         if(a[i][j]&&!a[i-1][j])

 23         {

 24             int p = i-1;

 25             while(p>=1&&!a[p][j])

 26             p--;

 27             a[p+1][j] = a[i][j];

 28             a[i][j] = 0;

 29         }

 30     }

 31     for(i = 2; i <= 10 ; i++)

 32     {

 33         if(a[1][i]&&!a[1][i-1])

 34         {

 35             int p = i-1;

 36             while(p>=1&&!a[1][p])

 37             p--;

 38             int tt = 1;

 39             while(tt<=12&&a[tt][i])

 40             {

 41                 a[tt][p+1] = a[tt][i];

 42                 a[tt][i] = 0;

 43                 tt+=1;

 44             }

 45         }

 46     }

 47 }

 48 void bfs(int x,int y)

 49 {

 50     if(!a[x][y]) return ;

 51     memset(vis,0,sizeof(vis));

 52     node te,tt;

 53     te.x = x;

 54     te.y = y;

 55     queue<node>q;

 56     q.push(te);

 57     int num = 1;

 58     vis[x][y] = 1;

 59     while(!q.empty())

 60     {

 61         tt = q.front();

 62         q.pop();

 63         for(int i = 0 ; i < 4 ; i++)

 64         {

 65             int tx = tt.x+dis[i][0];

 66             int ty = tt.y+dis[i][1];

 67             if(tx>=1&&tx<=12&&ty>=1&&ty<=10&&!vis[tx][ty]&&a[tx][ty]==a[x][y])

 68             {

 69                 vis[tx][ty] = 1;

 70                 te.x = tx;

 71                 te.y = ty;

 72                 num++;

 73                 q.push(te);

 74             }

 75         }

 76     }

 77     int i,j;

 78     if(num>=3)

 79     {

 80         for(i = 1; i <= 12 ; i++)

 81         for(j = 1; j <= 10 ; j++)

 82         if(vis[i][j])

 83         a[i][j] = 0;

 84     }

 85     init();

 86 }

 87 int main()

 88 {

 89     int n,i,j,u;

 90     char c;

 91     while(cin>>n)

 92     {

 93         if(!n) break;

 94         for(i = 12; i >= 1 ; i--)

 95             for(j = 1 ;j <= 10 ; j++)

 96             {

 97                 cin>>c;

 98                 a[i][j] = c-'A'+1;

 99             }

100         init();

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

102         {

103             cin>>c>>u;

104             int v = c-'a'+1;

105             bfs(u,v);

106         }

107         int ans = 0;

108         for(i = 1; i <= 12 ; i++)

109             for(j = 1; j <= 10 ; j++)

110             if(a[i][j])

111             ans++;

112         cout<<ans<<endl;

113     }

114     return 0;

115 }

116  

117 

118 

119 

120 /**************************************

121     Problem id    : SDUT OJ 2721 

122     User name    : shang 

123     Result        : Accepted 

124     Take Memory    : 460K 

125     Take Time    : 0MS 

126     Submit Time    : 2013-12-01 12:50:28  

127 **************************************/

View Code

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2723

枚举每条边 所存在除这条边以外的可以连接的路 就去掉这条边 水题

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<queue>

  7 #include<vector>

  8 #include<map>

  9 #include<string>

 10 using namespace std;

 11 char s1[210],s2[210],s[210][210];

 12 map<string,int>f;

 13 vector<int>ed[210];

 14 int w[210][210],vis[210],g;

 15 struct node

 16 {

 17     char s1[210],s2[210];

 18 }pp[40010],p[40010];

 19 int dfs(int u,int v)

 20 {

 21     queue<int>q;

 22     q.push(u);

 23     vis[u] = 1;

 24     while(!q.empty())

 25     {

 26         int k = q.front();

 27         q.pop();

 28         if(k==v) return 1;

 29         for(int i = 0; i < (int)ed[k].size() ; i++)

 30         {

 31             if(w[k][ed[k][i]]&&!vis[ed[k][i]])

 32             {

 33                 vis[ed[k][i]] = 1;

 34                 q.push(ed[k][i]);

 35             }

 36         }

 37     }

 38     return 0;

 39 }

 40 bool cmp(node a,node b)

 41 {

 42     if(strcmp(a.s1,b.s1)==0)

 43     return strcmp(a.s2,b.s2)<0;

 44     return strcmp(a.s1,b.s1)<0;

 45 }

 46 int main()

 47 {

 48     int n,kk=0,i,j;

 49     while(cin>>n)

 50     {

 51         if(!n) break;

 52         memset(w,0,sizeof(w));

 53         for(i = 0 ; i <= 200 ; i++)

 54         ed[i].clear();

 55         kk++;

 56         f.clear();

 57         g = 0;

 58         int oo=0;

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

 60         {

 61             cin>>s1>>s2;

 62             if(!f[s1])

 63             {

 64                 f[s1] = ++g;

 65                 strcpy(s[g],s1);

 66             }

 67             if(!f[s2]) {f[s2] = ++g;strcpy(s[g],s2);}

 68             w[f[s1]][f[s2]] = 1;

 69             ed[f[s1]].push_back(f[s2]);

 70         }

 71         int o = 0;

 72         for(i = 1; i <= g ; i++)

 73         {

 74             for(j = 0 ; j< (int)ed[i].size() ; j++)

 75             {

 76                 int v = ed[i][j];

 77                 w[i][v] = 0;

 78                 memset(vis,0,sizeof(vis));

 79                 if(dfs(i,v))

 80                 {

 81                     o++;

 82                     strcpy(p[o].s1,s[i]);

 83                     strcpy(p[o].s2,s[v]);

 84                 }

 85                 else

 86                 w[i][v] = 1;

 87             }

 88         }

 89         sort(p+1,p+o+1,cmp);

 90         printf("Case %d: ",kk);

 91         if(o!=0)

 92         {

 93             oo++;

 94             pp[oo] = p[1];

 95         }

 96         for(i = 2 ; i <= o ; i++)

 97         {

 98             if(strcmp(p[i].s1,p[i-1].s1)==0&&strcmp(p[i].s2,p[i-1].s2)==0)

 99             continue;

100             oo++;

101             pp[oo] = p[i];

102         }

103         printf("%d",oo);

104         for(i = 1 ; i < oo ; i++)

105         printf(" %s,%s",pp[i].s1,pp[i].s2);

106         if(oo!=0)

107         printf(" %s,%s",pp[oo].s1,pp[oo].s2);

108         puts("");

109     }

110     return 0;

111 }

112  

113 

114 

115 

116 /**************************************

117     Problem id    : SDUT OJ 2723 

118     User name    : shang 

119     Result        : Accepted 

120     Take Memory    : 17068K 

121     Take Time    : 520MS 

122     Submit Time    : 2013-12-01 14:12:31  

123 **************************************/

View Code