PKU 2481 - Cows (树状数组)

题意

输出满足题目要求的牛的数量。

思路

可以想到是树状数组。

其实第三个要求只要区间不是完全重合默认就满足了。我还想了很久。。。

先按y的大小排序,这样问题就变成统计比X小有多少个牛。处理一下完全相同的区间即可。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
struct POINT
{
    int x, y, id;
    bool operator < (const POINT &a) const
    {
        if (y != a.y) return y > a.y;
        return x < a.x;
    }
}pit[MAXN];
 
int C[MAXN + 10], ans[MAXN];
 
void Update(int k)
{
    while (k <= MAXN)
    {
        C[k]++;
        k += Lowbit(k);
    }
}
 
int Sum(int k)
{
    int ret = 0;
    while (k > 0)
    {
        ret += C[k];
        k -= Lowbit(k);
    }
    return ret;
}
 
int main()
{
    //ROP;
    int n, i, j;
    while (scanf("%d", &n), n)
    {
        MS(C, 0);
        for (i = 0; i < n; i++)
        {
            scanf("%d%d", &pit[i].x, &pit[i].y);
            pit[i].id = i;
            pit[i].x++;
        }
        sort(pit, pit + n);
        ans[pit[0].id] = 0;
        Update(pit[0].x);
        for (i = 1; i < n; i++)
        {
            if (pit[i].x == pit[i - 1].x && pit[i].y == pit[i - 1].y) ans[pit[i].id] = ans[pit[i - 1].id];
            else ans[pit[i].id] = Sum(pit[i].x);
            Update(pit[i].x);
        }
        for (i = 0; i < n; i++)
            if (i) printf(" %d", ans[i]);
            else printf("%d", ans[i]);
        puts("");
    }
    return 0;
}

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