【日常训练】【CF】20200604_方格_CF627E_two pointers/链表

题面

题解

现场得分：100/100

这道题我当场写了一个\(O(Rnk\log n)\)的东西。过了。

事情是如果枚举一条边界之后，倒序处理就能用链表，但是我是正序处理的，就只能用set。

下发题解

• 应该是dwh写的

代码

我的卡常代码

#include<bits/stdc++.h>
#define LL long long
#define MAXN 3010
using namespace std;
template<typename T>void Read(T &cn)
{
	char c; int sig = 1;
	while(!isdigit(c = getchar())) if(c == '-') sig = -1; cn = c-48;
	while(isdigit(c = getchar())) cn = cn*10+c-48; cn*=sig;
}
template<typename T>void Write(T cn)
{
	if(cn < 0) {putchar('-'); cn = 0-cn; }
	int wei = 0; T cm = 0; int cx = cn%10; cn/=10;
	while(cn) wei++, cm = cm*10+cn%10, cn/=10;
	while(wei--) putchar(cm%10+48), cm /= 10;
	putchar(cx+48);
}
template<typename T>void Max(T &cn, T cm) {cn = cn < cm ? cm : cn; }
template<typename T>void Min(T &cn, T cm) {cn = cn < cm ? cn : cm; }
struct qwe{
	int x,y;
	void getit() {Read(x); Read(y); }
	inline friend bool operator <(qwe a, qwe b) {return a.x == b.x ? a.y < b.y : a.x < b.x; }
};
struct qwer{
	int wei, num;
	void mk(int cn, int cm) {wei = cn; num = cm; }
	inline friend bool operator <(qwer a, qwer b) {return a.wei == b.wei ? a.num < b.num : a.wei < b.wei; }
};
int R,C,n,k;
qwe a[MAXN+1];
int wei1[12], wei2[12];
int wlen1, wlen2;
set<qwer> S;
typedef set<qwer>::iterator IT;
LL ans, zong;
void zeng(int cn, int cm)
{
//	printf("in zeng : cn = %d cm = %d\n",cn,cm);
	qwer lin; lin.mk(cn, cm);
	IT lin1 = S.upper_bound(lin);
	wlen1 = wlen2 = 0;
	for(IT i = lin1;i != S.end() && wlen2 < k;i++) wei2[++wlen2] = i->wei;
	if(lin1 != S.begin()) {
		lin1--; 
		for(IT i = lin1;i != S.begin() && wlen1 < k;i--) wei1[++wlen1] = i->wei;
		if(wlen1 < k && S.begin() != S.end()) wei1[++wlen1] = (S.begin())->wei;
	}
	if(wlen1 < k) wei1[++wlen1] = 0;
	if(wlen2 < k) wei2[++wlen2] = C+1;
//	for(int i = 0;i<=wlen1;i++) printf("%d ",wei1[i]); puts("");
//	for(int i = 0;i<=wlen2;i++) printf("%d ",wei2[i]); puts("");
	wei1[0] = wei2[0] = cn;
	for(int i = k-wlen1+1;i<=wlen2;i++) zong = 1ll*(wei2[i]-wei2[i-1])*(wei1[k-i]-wei1[k+1-i]) + zong;
	S.insert(lin);
}
int main()
{
//	freopen("rectangle.in","r",stdin);
//	freopen("rectangle.out","w",stdout);
	Read(R); Read(C); Read(n); Read(k);
	for(int i = 1;i<=n;i++) a[i].getit();
	sort(a+1, a+n+1); ans = 0;
	LL lsta = 0, lstx = -1;
	for(int i = 1;i<=R;i++)
	{
		int xian = 0; 
		while(xian <= n && a[xian+1].x < i) xian++;
		if(lstx == xian) {ans = ans + lsta; continue; }
		lsta = 0; lstx = xian; 
		zong = 0; S.clear();
		for(int j = i;j<=R;j++)
		{
			while(xian < n && a[xian+1].x <= j) {xian++; zeng(a[xian].y, xian); }
			lsta = lsta + zong;
		}
		ans = ans + lsta;
	}
	Write(ans); puts("");
	return 0;
}

题解代码

# include <bits/stdc++.h>
using namespace std;
namespace Base{
	# define mr make_pair
	typedef long long ll;
	typedef double db;
	const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
	const ll  infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
	template<typename T> void read(T &x){
    	x = 0; int fh = 1; double num = 1.0; char ch = getchar();
		while (!isdigit(ch)){ if (ch == '-') fh = -1; ch = getchar(); }
		while (isdigit(ch)){ x = x * 10 + ch - '0'; ch = getchar(); }
	    if (ch == '.'){
	    	ch = getchar();
	    	while (isdigit(ch)){num /= 10; x = x + num * (ch - '0'); ch = getchar();}
		}
		x = x * fh;
	}
	template<typename T> void chmax(T &x, T y){x = x < y ? y : x;}
	template<typename T> void chmin(T &x, T y){x = x > y ? y : x;}
}
using namespace Base;

const int N = 3010;
struct Node{
	int x, y;
}p[N];
int L, C, n, k, cnt[N], nxt[N], pre[N], num[N], les[N], hav[N];
ll ans;
bool cmp(Node x, Node y){
	return x.x < y.x || (x.x == y.x && x.y < y.y); 
}
void recal(int x){
	num[x] = (x - pre[x] - 1) * (x - pre[x]) / 2;
	int w = (x - pre[x]);
	while (les[x] != C + 1 && hav[x] < k){
		les[x] = nxt[les[x]];
		hav[x] += cnt[les[x]];
	}
	num[x] = num[x] + w * (les[x] - x);
}
int main(){
	freopen("rectangle.in", "r", stdin);
	freopen("rectangle.out", "w", stdout);
	read(L); read(C); read(n); read(k);
	for (int i = 1; i <= n; i++){
		read(p[i].x); read(p[i].y);
	}
	sort(p + 1, p + n + 1, cmp);
	for (int i = 1; i <= L; i++){
		memset(cnt, 0, sizeof(cnt));
		for (int j = n; j >= 1; j--){
			if (p[j].x < i) break;
			cnt[p[j].y]++;
		}
		int nt = C + 1;
		for (int j = C; j >= 1; j--)
			if (cnt[j] > 0) nxt[j] = nt, pre[nt] = j, nt = j;
		pre[nt] = 0; nxt[0] = nt; nxt[C + 1] = C + 2;
		int tot = C * (C + 1) / 2, non = 0;
		for (int j = nxt[0]; j != C + 2; j = nxt[j]){
			les[j] = j, hav[j] = cnt[j];
			recal(j);
			non = non + num[j];
		}
		ans = ans + tot - non;
		for (int t = L - 1, j = n; t >= i; t--){
			while (p[j].x > t){
				--cnt[p[j].y];
				int q = p[j].y, l = cnt[p[j].y];
				while (q != 0 && les[q] >= p[j].y){
					non = non - num[q];
					hav[q]--;
					recal(q);
					non = non + num[q];
					q = pre[q]; l = l + cnt[q];
				}
				if (cnt[p[j].y] == 0){
					nxt[pre[p[j].y]] = nxt[p[j].y];
					pre[nxt[p[j].y]] = pre[p[j].y];
					recal(nxt[p[j].y]);
				}
				j--;
			}
			ans = ans + tot - non;
		}
	}
	printf("%lld\n", ans);
	return 0;
}