2012暑假集训内部测试赛3

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2427

线段树+离散化  不离散化不知道会不会超时 一直RE 可能N值没有说的那么小吧 题意有问题 按1W开数组就RE 按10W开就A了

View Code

  1 #include<stdio.h>

  2 #include<string.h>

  3 #include<algorithm>

  4 #include<iostream>

  5 using namespace std;

  6 #define N 100001

  7 int s[N*6],num,f[100011],po[100011][2];

  8 struct node

  9 {

 10     int li,num;

 11 }line[200011];

 12 void build(int l,int r,int w)

 13 {

 14     s[w] = -1;

 15     if(l==r)

 16     {

 17         return ;

 18     }

 19     int m = (l+r)/2;

 20     build(l,m,2*w);

 21     build(m+1,r,2*w+1);

 22 }

 23 void add(int a,int b,int da,int l,int r,int w)

 24 {

 25     if(l==a&&b==r)

 26     {

 27         s[w] = da;

 28         return ;

 29     }

 30     int m = (l+r)/2;

 31     if(s[w]>0)

 32     {

 33         s[2*w] = s[w];

 34         s[2*w+1] = s[w];

 35         s[w] = -1;

 36     }

 37     if(b<=m)

 38         add(a,b,da,l,m,2*w);

 39     else

 40         if(a>m)

 41             add(a,b,da,m+1,r,2*w+1);

 42         else

 43         {

 44             add(a,m,da,l,m,2*w);

 45             add(m+1,b,da,m+1,r,2*w+1);

 46         }

 47 }

 48 void search(int l,int r,int w)

 49 {

 50     if(s[w]>0)

 51     {

 52         f[s[w]] = 1;

 53         return ;

 54     }

 55     if(l==r)

 56         return ;

 57     int m = (l+r)/2;

 58     search(l,m,2*w);

 59     search(m+1,r,2*w+1);    

 60 }

 61 bool cmp(node a,node b)

 62 {

 63     return a.num<b.num;

 64 }

 65 int main()

 66 {

 67     int n,i,j,t,a,b;

 68     scanf("%d", &t);

 69     while(t--)

 70     {

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

 72         num = 0;        

 73         scanf("%d", &n);

 74         build(1,N,1);    

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

 76         {

 77             scanf("%d%d", &po[i][0],&po[i][1]);

 78             line[2*i].li = i+1;

 79             line[2*i].num = po[i][0];

 80             line[2*i+1].li = -(i+1);

 81             line[2*i+1].num = po[i][1];

 82         }

 83         sort(line,line+2*n,cmp);    

 84         int te = line[0].num,g = 1;

 85         for(i = 0 ; i < 2*n ; i++)

 86         {

 87             if(te!=line[i].num)

 88             {

 89                 te = line[i].num;

 90                 g++;

 91             }

 92             if(line[i].li>0)

 93             {

 94                 po[line[i].li][0]=g;

 95             }

 96             else

 97                 po[-line[i].li][1]=g;

 98         }

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

100         {

101             add(po[i][0],po[i][1],i,1,N,1);

102         }

103         search(1,N,1);

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

105             if(f[i])



106                 num++;

107         printf("%d\n",num);

108     }

109     return 0;

110 } 

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2430

DP 前一个的1或者最高 到这一个1或者最高 中间选一个最优的

View Code

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 #include<cmath>

 5 using namespace std;

 6 int main()

 7 {

 8     int i,j,n,m,x,a[111],k;

 9     double dp[111][111];

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

11     {

12         memset(dp,0,sizeof(dp));

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

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

15         scanf("%d", &k);

16         if(n==1)

17         {

18             printf("0.000000\n");

19             continue;

20         }

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

22         {

23             x = a[i-1];

24             if(dp[i][1]<dp[i-1][x]+sqrt(k*k+(x-1)*(x-1)))

25             dp[i][1] = dp[i-1][x]+sqrt(k*k+(x-1)*(x-1));

26             if(dp[i][1]<dp[i-1][1]+k)

27             dp[i][1] = dp[i-1][1]+k;

28             if(dp[i][a[i]]<dp[i-1][x]+sqrt(k*k+(x-a[i])*(x-a[i])))

29             dp[i][a[i]] = dp[i-1][x]+sqrt(k*k+(x-a[i])*(x-a[i]));

30             if(dp[i][a[i]]<dp[i-1][1]+sqrt(k*k+(a[i]-1)*(a[i]-1)))

31             dp[i][a[i]] = dp[i-1][1]+sqrt(k*k+(a[i]-1)*(a[i]-1));

32         }

33         if(dp[n][1]>dp[n][a[n]])

34         printf("%.6lf\n",dp[n][1]);

35         else

36         printf("%.6lf\n",dp[n][a[n]]);

37     }

38     return 0;

39 }

40  

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2429

模拟 这题WA惨了 最后两分钟交对 好险。。

x不能为负值 就算最后有符合的负值也是输出-1 考虑两种情况x为0或者不为0的情况

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 int main()

 4 {

 5     int i,j,n,f[401],m;

 6     long long  k,a[61],stack[401],b[61];

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

 8     {

 9         int top =200;

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

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

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

13         {

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

15             b[i] = a[i];

16         }

17         scanf("%lld", &k);

18         for(j = 1; j <= n ; j++)

19         {

20             if(a[j]==-1)

21                 a[j]=0;

22             if(a[j]==0)

23             {

24                 top--;

25                 long long  x = stack[top];

26                 top--;

27                 x+=stack[top];

28                 stack[top++] = x;

29             }

30             else

31                 stack[top++] = a[j];

32         }

33         top--;

34         if(stack[top]==k)

35         {

36             printf("0\n");

37             continue;

38         }

39         top = 200;

40         for(j = 1; j <= n ; j++)

41         {

42             if(b[j]==-1)

43             {

44                 f[top]=1;

45                 stack[top++] = 0;

46             }

47             else

48             if(b[j]==0)

49             {

50                 top--;

51                 long long  x = stack[top];

52                 top--;

53                 x+=stack[top];

54                 if(f[top]||f[top+1])

55                 {

56                     f[top] = 1;

57                 }

58                 stack[top++] = x;        

59             }

60             else

61                 stack[top++] = b[j];

62         }

63         top--;

64         if(!f[top]&&stack[top]==k)

65         {

66             printf("0\n");

67             continue;

68         }

69         if(f[top]&&k-stack[top]>0)

70             printf("%lld\n",k-stack[top]);

71         else

72             printf("-1\n");

73     }

74     return 0;

75 }