POJ 2828 Buy Tickets(树状数组+二分)

题目链接

看了别人思路，逆序插入，然后就会发现只要为留出pos[i]个位置就行了。树状数组+二分求下界就好了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define N 200011
int pos[N];
int p[N],n;
struct node
{
    int val;
    int id;
}num[N];
int lowbit(int t)
{
    return t&(-t);
}
void insert(int t,int d)
{
    while(t <= n)
    {
        p[t] += d;
        t += lowbit(t);
    }
}
int getsum(int t)
{
    int sum = 0;
    while(t > 0)
    {
        sum += p[t];
        t -= lowbit(t);
    }
    return sum;
}
int cmp(node a,node b)
{
    return a.id < b.id;
}
int main()
{
    int i,str,end,temp,mid;
    while(scanf("%d",&n)!=EOF)
    {
        for(i = 0;i < n;i ++)
        {
            scanf("%d%d",&pos[i],&num[i].val);
            p[i] = 0;
        }
        for(i = 0;i < n;i ++)
        {
            insert(i+1,1);
        }
        for(i = n-1;i >= 0;i --)
        {
            str = 1;
            end = n;
            while(str < end)
            {
                mid = (str+end)/2;
                temp = getsum(mid);
                if(temp < pos[i]+1)//求上界
                str = mid + 1;
                else
                end = mid;
            }
            num[i].id = str;
            insert(str,-1);
        }
        sort(num,num+n,cmp);
        for(i = 0;i < n;i ++)
        {
            if(i == 0)
            printf("%d",num[i].val);
            else
            printf(" %d",num[i].val);
        }
        printf("\n");
    }
    return 0;
}