2012暑假集训内部测试赛1

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2417

找峰值点 从小到大输出

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 int a[50001],b[50001];

 4 int main()

 5 {

 6     int i,j,k,n,m;

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

 8     {

 9         int g = 0;

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

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

12         if(n==1)

13         {

14             printf("1\n");

15             continue;

16         }

17         if(a[1]>=a[2])

18             b[g++] = 1;

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

20         {

21             if(a[i-1]<=a[i]&&a[i]>=a[i+1])

22                 b[g++] = i;

23         }

24         if(a[n]>=a[n-1])

25             b[g++] = n;

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

27             printf("%d\n",b[i]);

28     }

29     return 0;

30 }

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2418

并查集 注意是多组

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<math.h>

 4 struct node

 5 {

 6     int u,v;

 7 }q[10011];

 8 int father[10011],r[10011],w[111][111];

 9 int find(int x)

10 {

11     if(x!=father[x])

12         father[x] = find(father[x]);

13     return father[x];

14 }

15 int main()

16 {

17     int n,m,k,i,j,a,b,g = 0;

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

19     {

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

21         g = 0;

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

23         {

24             father[i] = i;

25             r[i] = 1;

26         }

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

28         {

29             scanf("%d%d", &a,&b);

30             if(w[a][b]==0&&a>=1&&a<=n&&b>=1&&b<=m)

31             {

32                 q[++g].u = a;

33                 q[g].v = b;

34                 w[a][b]  =1;

35             }

36         }

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

38         {

39             for(j = i+1; j <= g ; j++)

40             {

41                 if((abs(q[i].u-q[j].u)==1&&q[i].v==q[j].v)||(q[i].u==q[j].u&&abs(q[i].v-q[j].v)==1))

42                 {

43                     int px = find(i);

44                     int py = find(j);

45                     if(px!=py)

46                     {

47                         father[px] = py;

48                         r[py]+=r[px];

49                     }

50                 }

51             }

52         }

53         int max = 0;

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

55             if(max<r[i])

56                 max= r[i];

57         printf("%d\n",max);

58     }

59     return 0;

60 }

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2420

记录每个1前面2的个数 和每个1后面1的个数 找min{d[i][1]+d[i][2]}

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 int a[30001],b[30001],d[30001][3];

 4 int main()

 5 {

 6     int i,j,k,n,m,min;

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

 8     {

 9         memset(d,0,sizeof(d));

10         min = 50000;

11         int f = 0;

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

13         {

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

15         }

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

17         {

18             if(a[i]==1)

19                 d[i][2] = d[i-1][2];

20             else

21                 d[i][2] = d[i-1][2]+1;

22         }

23         for(i = n ; i >= 1; i--)

24         {

25             if(a[i]==2)

26                 d[i][1] = d[i+1][1];

27             else

28                 d[i][1] = d[i+1][1]+1;

29         }

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

31         {

32             if(a[i]==1)

33             {

34                 if(min>d[i][1]+d[i][2]-1)

35                 min = d[i][1]+d[i][2]-1;

36                 f = 1;

37             }

38         }

39         if(!f)

40             min = 0;

41         printf("%d\n",min);

42     }

43     return 0;

44 }

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2421

这题描述的太不清楚了 n值并不表示 最多就有n个点 从输入里面找一个最大值

因为走到头要返回 而只有最后一次走不用返回 所以最后走那一条最长的

最小生成树 求出和S 用S-生成树中最长的那一条

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 #define INF 100000000

 6 int w[501][501],d[501],vis[501],dis[501];

 7 void prime(int n)

 8 {

 9     int i,j,k,s = 0;

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

11     vis[0] = 1;

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

13     {

14         d[i] = w[0][i];

15         if(w[0][i]!=INF)

16         dis[i] = w[0][i];

17         else

18         dis[i]= 0;

19     }

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

21     {

22         int min = INF;

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

24         if(!vis[j]&&d[j]<min)

25         min = d[k= j];

26         if(min == INF)

27         break;

28         s+=min;

29         vis[k] = 1;

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

31         if(!vis[j]&&d[j]>w[k][j])

32         {

33             d[j] = w[k][j];

34             dis[j] = dis[k]+w[k][j];

35         }

36     }

37     int mi = -1;

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

39     {

40 

41         if(dis[i]>mi)

42         mi = dis[i];

43     }

44     printf("%d\n",s*2-mi);

45 }

46 int main()

47 {

48     int i,j,k,n,m,a,b,c;

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

50     {

51        for(i = 0 ;i <= 100; i++)

52        {

53           for(j = 0 ; j <= 100; j++)

54           w[i][j] = INF;

55           w[i][i] = 0;

56        }

57         m = 0;

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

59         {

60             scanf("%d%d%d",&a,&b,&c);

61             if(w[a][b]>c)

62             {

63                 w[a][b] = c;

64                 w[b][a] = c;

65             }

66             if(a>m)

67             m = a;

68             if(b>m)

69             m = b;

70         }

71         prime(m);

72     }

73     return 0;

74 }