2010山东省第一届ACM程序设计竞赛

休眠了2月了 要振作起来了！！。。。

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2155

因为点比较少 最多更新三百次 标记某个节点时直接更新与之相连的点的最短距离

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 using namespace std;

 8 #define INF 0xfffffff

 9 int w[310][310],f[310];

10 int main()

11 {

12     int n,m,i,j,q;

13     int kk = 1;

14     while(cin>>n>>m>>q)

15     {

16         if(n==0&&m==0&&q==0) break;

17         int u,v,c;

18         memset(f,0,sizeof(f));

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

20         {

21             for(j = 0 ;j < n ; j++)

22             w[i][j] = INF;

23             w[i][i] = 0;

24         }

25         for(i = 1; i <= m ; i++)

26         {

27             cin>>u>>v>>c;

28             w[u][v] = min(w[u][v],c);

29         }

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

31         while(q--)

32         {

33             int t;

34             cin>>t;

35             if(t==1)

36             {

37                 int x,y;

38                 cin>>x>>y;

39                 if(!f[x]||!f[y])

40                 printf("City %d or %d is not available.\n",x,y);

41                 else if(w[x][y]==INF)

42                 puts("No such path.");

43                 else

44                 printf("%d\n",w[x][y]);

45             }

46             else

47             {

48                 int x;

49                 cin>>x;

50                 if(f[x])

51                 {

52                     printf("City %d is already recaptured.\n",x);

53                     continue;

54                 }

55                 f[x] = 1;

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

57                     for(j = 0 ; j < n ; j++)

58                     if(w[i][j]>w[i][x]+w[x][j])

59                     w[i][j] = w[i][x]+w[x][j];

60             }

61         }

62         puts("");

63     }

64     return 0;

65 }

66  

67 

68 

69 

70 /**************************************

71     Problem id    : SDUT OJ 2155 

72     User name    : shang 

73     Result        : Accepted 

74     Take Memory    : 832K 

75     Take Time    : 300MS 

76     Submit Time    : 2014-01-16 15:11:20  

77 **************************************/

View Code

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2159 

set存结构体 low_bounder二分查找 不过时间跑了不少 不知道有没有更简单的方法

 1 #include<cstdio>

 2 #include<cstring>

 3 #include <cmath>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<queue>

 7 #include<vector>

 8 #include<set>

 9 #include<iostream>

10 using namespace std;

11 struct node

12 {

13     int x,y;

14     bool operator <(const node a) const

15     {

16         if(a.x==x)

17         return y<a.y;

18         return x<a.x;

19     }

20 };

21 char s[10];

22 int main()

23 {

24     int n,kk=1;

25     while(cin>>n)

26     {

27         if(!n) break;

28         set<node>q;

29         node st;

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

31         while(n--)

32         {

33             cin>>s;

34             if(s[0]=='a')

35             {

36                 cin>>st.x>>st.y;

37                 q.insert(st);

38             }

39             else if(s[0]=='f')

40             {

41                 cin>>st.x>>st.y;

42                 set<node>::iterator it;

43                 it = q.lower_bound(st);

44                 while(it!=q.end())

45                 {

46                     if((*it).x>st.x&&(*it).y>st.y)

47                     {

48                         cout<<(*it).x<<" "<<(*it).y<<endl;

49                         break;

50                     }

51                     it++;

52                 }

53                 if(it==q.end())

54                 puts("-1");

55             }

56             else

57             {

58                 cin>>st.x>>st.y;

59                 q.erase(st);

60             }

61         }

62         puts("");

63     }

64     return 0;

65 }

66  

67 

68 

69 

70 /**************************************

71     Problem id    : SDUT OJ 2159 

72     User name    : shang 

73     Result        : Accepted 

74     Take Memory    : 2572K 

75     Take Time    : 930MS 

76     Submit Time    : 2014-01-16 17:34:20  

77 **************************************/

View Code

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2157

先取两个数 再二分查找和与m最近的数

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 using namespace std;

 8 int a[1010],p[1010*1010];

 9 int main()

10 {

11     int n,m,i,j;

12     int kk = 1;

13     while(scanf("%d%d",&n,&m)!=EOF)

14     {

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

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

17         scanf("%d",&a[i]);

18         a[n] = 0;

19         int g = 0;

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

21             for(j = i ; j <= n ; j++)

22             if(a[i]+a[j]<=m)

23             p[g++] = a[i]+a[j];

24         sort(p,p+g);

25         int maxz = 0;

26         printf("Case %d: ",kk++);

27         for(i = 0 ; i < g ;  i++)

28         {

29             if(p[i]>m) break;

30             int low = i,high = g-1;

31             while(low<high)

32             {

33                 int mm = (low+high)/2;

34                 if(p[i]+p[mm]>m)

35                 high = mm-1;

36                 else

37                 low = mm+1;

38             }

39             if(p[low]+p[i]<=m&&p[low]+p[i]>maxz)

40             {

41                 maxz = p[low]+p[i];

42             }

43         }

44         printf("%d\n",maxz);

45         puts("");

46     }

47     return 0;

48 }

49  

50 

51 

52 

53 /**************************************

54     Problem id    : SDUT OJ 2157 

55     User name    : shang 

56     Result        : Accepted 

57     Take Memory    : 2444K 

58     Take Time    : 130MS 

59     Submit Time    : 2014-01-16 15:43:45  

60 **************************************/

View Code