51Nod－1107－斜率小于0的连线数量

来源：互联网发布：学编程用什么软件编辑：程序博客网时间：2024/06/05 15:03

ACM模版

描述

题解

常规解法是逆序数（也可以使用树状数组解），注意横坐标相等的情况。

代码

One：

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 5e4 + 10;long long ans = 0;struct line{    int x;    int y;} Line[MAXN];line L[MAXN];bool cmp(line a, line b){    if (a.x == b.x) //  特殊考虑横坐标相等的情况    {        return a.y < b.y;    }    return a.x < b.x;}//  归并排序求逆序数void MergeSort(int l, int r){    int mid, i, j, tmp;    if (r > l + 1)    {        mid = (l + r) / 2;        MergeSort(l, mid);        MergeSort(mid, r);        tmp = l;        for (i = l, j = mid; i < mid && j < r;)        {            if (Line[i].y > Line[j].y)            {                L[tmp++] = Line[j++];                ans += mid - i;            }            else            {                L[tmp++] = Line[i++];            }        }        if (j < r)        {            for (; j < r; j++)            {                L[tmp++] = Line[j];            }        }        else        {            for (; i < mid; i++)            {                L[tmp++] = Line[i];            }        }        for (i = l; i < r; i++)        {            Line[i] = L[i];        }    }    return ;}int main(int argc, const char * argv[]){//    freopen("/Users/zyj/Desktop/input.txt", "r", stdin);    int N;    cin >> N;    for (int i = 0; i < N; i++)    {        scanf("%d %d", &Line[i].x, &Line[i].y);    }    sort(Line, Line + N, cmp);    MergeSort(0, N);    std::cout << ans << '\n';    return 0;}

Two：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 5e4 + 5;struct node{    int x, y;    inline bool operator < (const node &nd)    {        if (this->x == nd.x)        {            return this->y > nd.y;        }        return this->x > nd.x;    }} nums[MAXN];int bit[MAXN];inline bool cmp(const node &nd1, const node &nd2){    return nd1.y < nd2.y;}inline int sum(int i, int *bit){    int res = 0;    for (; i > 0; i ^= i & -i)    {        res += bit[i];    }    return res;}inline void add(int i, int adder, int *bit){    for (; i < 50001; i += i & -i)    {        bit[i] += adder;    }}int main(){    int n, res = 0, last;    scanf("%d", &n);    for (int i = 0; i < n; i++)    {        scanf("%d%d", &nums[i].x, &nums[i].y);    }    //  坐标离散化    sort(nums, nums + n, cmp);    last = nums[0].y;    nums[0].y = 0;    for (int i = 1; i < n; i++)    {        int tmp = nums[i].y;        nums[i].y = nums[i - 1].y + (last < nums[i].y);        last = tmp;    }    //  对x排序，准备计算    sort(nums, nums + n);    for (int i = 0; i < n; i++)    {        //  计算目前比y小的点        res += sum(nums[i].y, bit);        //  更新bit        add(nums[i].y + 1, 1, bit);    }    printf("%d", res);    return 0;}

参考

《逆序数》

0 0