2016 UESTC Training for Data Structures E - 卿学姐与城堡的墙 树状数组求逆序对、离散化

E - 卿学姐与城堡的墙

Input

Output

Sample input and output

3 -3 1
-1 3
2 2
1 1

3

Hint

Source

My Solution

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;
const int maxn = 200000 + 8;

map<long long,int> s;      //用来处理 u == v 的情况
int N;               //具体操作的时候要用所以放到全局
struct segment{
long long vy, uy;
int order;
} seg[maxn];

bool cmpu(const segment& a, const segment& b)
{
if(a.uy != b.uy) return a.uy > b.uy;
else return a.vy < b.vy;
}

bool cmpv(const segment& a, const segment& b)
{
if(a.vy != b.vy) return a.vy > b.vy;
else return a.uy < b.uy;
}

//!树状数组求逆序对
int Tree[maxn];

inline int lowbit(int x)
{
return (x&-x);
}

void add(int x, int value)
{
for(int i = x; i <= N; i += lowbit(i))
Tree[i] += value;
}

int get(int x)
{
int sum = 0;
for(int i = x; i; i -= lowbit(i))
sum += Tree[i];
return sum;
}

//!!!!!! u, v, k, b; 会在seg[i].uy = k*u + b; seg[i].vy = k*v + b;这中间过程溢出，所以也要 long long 才行了。
//!!!!!! 这里改了然后过来，WA了10次test1，......test就玩溢出，坑坏我了，前天被坑起啊(┬＿┬)
int main()
{
#ifdef LOCAL
freopen("a.txt", "r", stdin);
#endif // LOCAL
long long u, v, k, b;
//!u <= v   ...... u == v
long long ans = 0, crosspointsinu = 0;
scanf("%d%lld%lld", &N, &u, &v);
//!....... u == v 的情况处理错了

for(int i = 0; i < N; i++){
scanf("%lld%lld", &k, &b);
seg[i].uy = k*u + b;
seg[i].vy = k*v + b;
s[seg[i].uy]++;
}
//据说可以用set去重
sort(seg, seg + N, cmpu);
for(int i = 0; i < N; i++){
seg[i].order = i+1;   //根据u的值进行离散化
}
//

if(u == v){
int times;
for(auto i = s.begin(); i != s.end(); i++){
times = i->second;
if(times != 1){
crosspointsinu += (times*(times-1))/2;
}
}
printf("%lld", crosspointsinu);
}
else{
sort(seg, seg + N, cmpv);

memset(Tree, 0, sizeof Tree);
for(int i = 1; i <= N; i++){
add(seg[i-1].order, 1);
ans += i - get(seg[i-1].order);

}
printf("%lld", ans);
}
return 0;
}