Gym - 101853C Intersections

Gym - 101853C Intersections

传送门Gym - 101853C

题意

就是给你n，然后给你两个n的排列，让你把两个排列中相同的连线，求有多少对相交的线

题解

第一反应就是对于每个数来说，他在第一行的位置的左边的数，连到他在第二行的位置的右边的数的个数。然后对于每个数求个和。然后又想到对第一排重新标号，那就是对重标号完的第二行求逆序对。那就很好写了~只不过我爆int，wa了两发。

ac代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <iomanip>
//#include <unordered_map>

#pragma comment(linker, "/STACK:102400000,102400000")
#define fir first
#define sec second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define clr(x) memset(x,0,sizeof(x))
#define cld(x) memset(x,-1,sizeof(x))
#define clx(x) memset(x,63,sizeof(x))
#define cln(x) memset(x,-64,sizeof(x))
#define rush() int T;scanf("%d",&T);while(T--)
#define pi 3.1415926
#define VM 100047
#define EM 400047
#define rd(x) scanf("%d",&x);

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const ll llf = 0x3f3f3f3f3f3f3f3f;
const int maxn = (int) 1e6 + 7;
const double eps = 1e-10;
const ll mod1 = (int) 1e9 + 7;
const ll mod2 = 998244353;
const ll has = 99959;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};


int a[maxn], c[maxn], n;
map<int, int> mp;
ll ans;

void x(int l, int r) {
    int mid = (l + r) / 2, i, j, tmp;
    if (r > l) {
        x(l, mid);
        x(mid + 1, r);
        tmp = l;
        for (i = l, j = mid + 1; i <= mid && j <= r;) {
            if (a[i] > a[j]) {
                c[tmp++] = a[j++];
                ans += mid - i + 1;
            } else c[tmp++] = a[i++];
        }
        if (i <= mid) for (; i <= mid;) c[tmp++] = a[i++];
        if (j <= r) for (; j <= r;) c[tmp++] = a[j++];
        for (i = l; i <= r; i++) a[i] = c[i];
    }
}

int main() {
    std::ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    rush() {
        mp.clear();
        ans = 0;
        clr(a);
        clr(c);
        scanf("%d", &n);
        int xx;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &xx);
            mp[xx] = i;
        }
        for (int i = 1; i <= n; i++) {
            scanf("%d", &xx);
            a[i] = mp[xx];
        }
        x(1, n);
        printf("%lld\n", ans);
    }
    return 0;
}