POJ 2481 Cows

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解题思路：

实质是排序后求逆序对

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = 100000 + 10;
struct Node
{
    int l, r;
    int id;
    bool operator < (const Node& rhs)const
    {
        if(l != rhs.l)
            return l < rhs.l;
        else
            return r > rhs.r;
    }
}cow[MAXN];
struct Tree
{
    int l, r;
    int sum;
}tree[MAXN<<2];
int ans[MAXN];
void build(int l, int r, int rt)
{
    tree[rt].l = l;
    tree[rt].r = r;
    tree[rt].sum = 0;
    if(l == r) return ;
    int m = (l + r) >> 1;
    build(l, m, rt<<1);
    build(m + 1, r, rt<<1|1);
}
void update(int p, int rt)
{
    tree[rt].sum ++;
    if(tree[rt].l == tree[rt].r) return ;
    else
    {
        int m = (tree[rt].l + tree[rt].r) >> 1;
        if(p <= m) update(p, rt<<1);
        else update(p, rt<<1|1);
    }
}
int Query(int l, int r, int rt)
{
    if(tree[rt].l >= l && tree[rt].r <= r)
        return tree[rt].sum;
    else
    {
        int m = (tree[rt].l + tree[rt].r) >> 1;
        int sum = 0;
        if(l <= m) sum += Query(l, r, rt<<1);
        if(r > m) sum += Query(l, r, rt<<1|1);
        return sum;
    }
}
int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        if(n == 0)
            break;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d", &cow[i].l, &cow[i].r);
            cow[i].id = i;
        }
        sort(cow, cow + n);
        build(0, MAXN, 1);
        for(int i=0;i<n;i++)
        {
            if(i != 0 && cow[i].l == cow[i-1].l && cow[i].r == cow[i-1].r)
                ans[cow[i].id] = ans[cow[i-1].id];
            else
                ans[cow[i].id] = Query(cow[i].r, MAXN, 1);
            update(cow[i].r, 1);
        }
        for(int i=0;i<n-1;i++)
            printf("%d ", ans[i]);
        printf("%d\n", ans[n-1]);
    }
    return 0;
}

POJ 2481 Cows

标签：

原文地址：http://blog.csdn.net/moguxiaozhe/article/details/45196291