51Nod－1107－斜率小于0的连线数量

描述

题解

代码

#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;
}

#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