• poj 2182 Lost Cows 解题报告


    题意:每个奶牛都有一个编号,1- N 从第二个牛开始给出前面比她编号小的牛的个数,问你求牛的编号序列

    解题思路:线段树+ 二分查找 (多个相同的数二分边界问题需要注意) 

    解题代码:

      1 #include <stdlib.h>
      2 #include <string.h>
      3 #include <stdio.h>
      4 #define MAXN 8005
      5 struct node
      6 {
      7   int left , right,mid ;
      8   int num;
      9 }tree[MAXN*4];
     10 int  L(int c)
     11 {
     12   return 2 * c;
     13 }
     14 int R(int c)
     15 {
     16   return 2 * c  + 1;
     17 }
     18 void up(int c )
     19 {
     20   tree[c].num = tree[L(c)].num + tree[R(c)].num;
     21 }
     22 void  build(int c ,int p , int v)
     23 {
     24    tree[c].left = p ;
     25    tree[c].right = v ;
     26    tree[c].mid = (p+v)/2;
     27    tree[c].num = 1;
     28    if(p == v )
     29    {
     30      return;
     31    }
     32    build(L(c),p,tree[c].mid);
     33    build(R(c),tree[c].mid + 1, v );
     34    up(c);
     35 }
     36 void update(int c , int p)
     37 {
     38    if(tree[c].left == p && tree[c].right == p )
     39    {
     40        tree[c].num = 0 ;
     41        return ;
     42    }
     43    if(p <= tree[c].mid) update(L(c),p);
     44    else update(R(c),p);
     45    up(c);
     46 }
     47 int tsum = 0 ;
     48 void getsum (int c, int p , int v )
     49 {
     50    if(p <= tree[c].left && v >= tree[c].right)
     51    {
     52      tsum += tree[c].num;
     53      return ;
     54    }
     55    if(v <= tree[c].mid) getsum (L(c),p,v);
     56    else if(p > tree[c].mid) getsum(R(c),p, v);
     57    else
     58    {
     59       getsum(L(c),p,tree[c].mid);
     60      getsum(R(c),tree[c].mid + 1, v );
     61    }
     62 }
     63 int a[MAXN];
     64 int b[MAXN];
     65 int main()
     66 {
     67    int n ;
     68    while(scanf("%d",&n) != EOF)
     69    {
     70      memset(a,0,sizeof(a));
     71      memset(b,0,sizeof(b));
     72      for(int i = 2; i <= n;i ++)
     73         scanf("%d",&a[i]);
     74      b[n] = a[n] + 1;
     75      build(1,1,n+1);
     76      update(1,b[n]);
     77      for(int i = n- 1; i >=1 ;i --)
     78      {
     79        int low = 1 , high = n;
     80        int ans ;
     81        while(low <= high)
     82        {
     83          tsum = 0 ;
     84 
     85          int mid = (low + high)/2;
     86          getsum(1,1,mid);
     87          if(tsum >= a[i]+1)
     88          {
     89            ans = mid ;
     90            high = mid - 1;
     91          }
     92          else
     93             low = mid +1;
     94        }
     95      //  printf("%d
    ",ans);
     96        b[i] = ans ;
     97        tsum = 0 ;
     98        getsum(1,1,ans);
     99      //  printf("%d
    ",tsum);
    100        update(1,b[i]);
    101      }
    102      for(int i = 1;i <= n; i ++)
    103         printf("%d
    ",b[i]);
    104    }
    105    return 0 ;
    106 }
    View Code
    没有梦想,何谈远方
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  • 原文地址:https://www.cnblogs.com/zyue/p/3224570.html
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