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\n",ans);
 96        b[i] = ans ;
 97        tsum = 0 ;
 98        getsum(1,1,ans);
 99      //  printf("%d\n",tsum);
100        update(1,b[i]);
101      }
102      for(int i = 1;i <= n; i ++)
103         printf("%d\n",b[i]);
104    }
105    return 0 ;
106 }