2019 上海網絡賽 C. Triple （FFT + 暴力）

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題目大意：如果對$A_i ,B_j,C_k$排序，使得$A_i ,B_j,C_k$恰好是升序，要滿足的條件爲：$C_k - B_j \geq A_i$，這個定義類似於從$A,B,C$數組中各取一個數，使得能構成三角形，但是注意這裏是可以等於的。

對於這種題有一個FFT的做法：通過FFT求出 $a + b$的所有方案，然後枚舉第三條邊，扣掉不合法的方案。FFT複雜度與 $n$ 無關，與邊權最大值有關，跑100組 $10^5 \log 10^5$ 再加上FFT的大常數，肯定是要T的。

題目說到只有20組 $n > 1000$，對於 $n < 1000$的情況，直接用 $n ^ 2$暴力方法即可

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代碼：

#include<iostream>
using namespace std;
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int mx = 1e5;
typedef long long ll;
struct complex{
	double r,i;
	complex(double _r = 0.0,double _i = 0.0) {
		r = _r;
		i = _i;
	}
	complex operator + (const complex & rhs) {
		return complex(rhs.r + r,rhs.i + i);
	}
	complex operator - (const complex & rhs) {
		return complex(r - rhs.r,i - rhs.i);
	}
	complex operator * (const complex & rhs) {
		return complex(r * rhs.r - i * rhs.i,r * rhs.i + i * rhs.r);
	}
};
complex A[maxn],B[maxn];
int a[4][maxn],n,t;
int cnt[4][maxn];
int sum[maxn];
void change(complex a[],int len) {
	for(int i = 1,j = len / 2; i < len - 1; i++) {
		if(i < j) swap(a[i],a[j]);
		int k = len / 2;
		while(j >= k) {
			j -= k;
			k /= 2;
		}
		j += k;
	} 
}
void fft(complex a[],int len,int type) {
	change(a,len);
	for(int i = 2; i <= len; i <<= 1) {
		complex wp = complex(cos(2 * pi * type / i),sin(2 * pi * type / i));
		for(int j = 0; j < len; j += i) {
			complex w = complex(1,0);
			for(int k = 0; k < i / 2; k++) {
				complex t = a[j + k];
				complex u = w * a[j + k + i / 2];
				a[j + k] = t + u;
				a[j + k + i / 2] = t - u;
				w = w * wp;
			}
		}
	}
	if(type == -1) {
		for(int i = 0; i < len; i++)
			a[i].r /= len;
	} 
}
ll cal() {
	memset(sum,0,sizeof sum);
	ll ans = 1ll * n * n * n;	
	for(int i = 1; i <= n; i++) 
		sum[a[3][i]]++;
	for(int i = 1; i <= mx; i++)
		sum[i] += sum[i - 1];
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n && a[2][j] < a[1][i]; j++) {
			ll tmp = a[1][i] - a[2][j] - 1;
			ans -= sum[tmp];
		}
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n && a[1][j] < a[2][i]; j++) {
			ll tmp = a[2][i] - a[1][j] - 1;
			ans -= sum[tmp];
		}
	}
	for(int i = 0; i <= mx; i++) sum[i] = 0;
	for(int i = 1; i <= n; i++)
		sum[a[2][i]]++;
	for(int i = 1; i <= mx; i++) 
		sum[i] += sum[i - 1];
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n && a[1][j] < a[3][i]; j++) {
			ll tmp = a[3][i] - a[1][j] - 1;
			ans -= sum[tmp];
		}
	}	
	return ans;
}
ll solve() {
	int len;
	ll ans = 1ll * n * n * n;
	len = 1;
	while(len <= 2 * a[1][n]) len <<= 1;
	for(int i = 0; i <= mx; i++) {
		A[i] = complex(cnt[2][i],0);
		B[i] = complex(cnt[3][i],0);
	}
	for(int i = mx + 1; i < len; i++) {
		A[i] = complex(0,0);
		B[i] = complex(0,0);
	}
	fft(A,len,1);fft(B,len,1);
	for(int i = 0; i < len; i++)
		A[i] = A[i] * B[i];
	fft(A,len,-1);
	for(int i = 1; i < len; i++)
		A[i] = A[i] + A[i - 1];
	for(int i = 1; i <= n; i++) {
		ans -= A[a[1][i] - 1].r;
	}
	len = 1;
	while(len <= 2 * a[2][n]) len <<= 1;
	for(int i = 0; i < len; i++) {
		if(i <= mx) A[i] = complex(cnt[1][i],0);
		else A[i] = complex(0,0);
	}	
	fft(A,len,1);
	for(int i = 0; i < len; i++)
		A[i] = A[i] * B[i];
	fft(A,len,-1);
	for(int i = 1; i < len; i++)
		A[i] = A[i] + A[i - 1];
	for(int i = 1; i <= n; i++) {
		ans -= A[a[2][i] - 1].r;
	}
	len = 1;
	while(len <= 2 * a[3][n]) len <<= 1;	
	for(int i = 0; i < len; i++) {
		if(i <= mx) {
			A[i] = complex(cnt[1][i],0);
			B[i] = complex(cnt[2][i],0);
		}
		else {
			A[i] = complex(0,0);
			B[i] = complex(0,0);
		}
	}		
	fft(A,len,1);fft(B,len,1);
	for(int i = 0; i < len; i++)
		A[i] = A[i] * B[i];
	fft(A,len,-1);
	for(int i = 1; i < len; i++)
		A[i] = A[i] + A[i - 1];
	for(int i = 1; i <= n; i++) {
		ans -= A[a[3][i] - 1].r;
	}
	return ans;
}
int main() {
	scanf("%d",&t);
	int ca = 0;
	while(t--) {
		scanf("%d",&n);
		for(int i = 1; i <= n; i++)
			scanf("%d",&a[1][i]);
		sort(a[1] + 1,a[1] + n + 1);
		for(int i = 1; i <= n; i++)
			scanf("%d",&a[2][i]);
		sort(a[2] + 1,a[2] + n + 1);
		for(int i = 1; i <= n; i++)
			scanf("%d",&a[3][i]);		
		sort(a[3] + 1,a[3] + n + 1);
		for(int i = 0; i <= mx; i++)
			cnt[1][i] = cnt[2][i] = cnt[3][i] = 0;
		for(int i = 1; i <= n; i++) {
			cnt[1][a[1][i]]++;
			cnt[2][a[2][i]]++;
			cnt[3][a[3][i]]++;
		}		
		if(n <= 2000)	 {
			printf("Case #%d: %lld\n",++ca,cal());
		}
		else {
			printf("Case #%d: %lld\n",++ca,solve());
		}
	}
	return 0;
}