屯题？

BZOJ1635

傻逼题，差分一下。

#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
const int MAXN = 10000;
using namespace std;

inline int read() {
	int x = 0, w = 1; char c = ' ';	
	while (c < '0' || c > '9') {c = getchar(); if (c == '-') w = -1;}
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
	return x * w;
}

map < pair <int , int>, bool > H;

int c[MAXN + 5], S[MAXN + 5];

int main() {
	int n = read(), p = read(), h = read(), r = read(), tot = 0;
	
	for (int i = 1; i <= r; ++i) {
		int x = read(), y = read();
		if (x > y) swap(x, y);
		
		if (H.find(make_pair(x, y))->second) continue;
		
		--c[x + 1], ++c[y];
		H[make_pair(x, y)] = true;
	}
	
	for (int i = 1; i <= n; ++i) {
		S[i] = S[i - 1] + c[i];
		printf("%d\n", h + S[i]);
	}
	
	return 0;
}

POJ2018

\(二分+DP\)，其实就是乱搞，用斜率优化的\(DP\)也可以做。挺裸的，就是精度有点毒。

#include <iostream>
#include <cstdio>
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a < b ? b : a)
const double INF = 1e10;
const int MAXN = 100000;
const double eps = 1e-5;
using namespace std;

int n, f;
double a[MAXN + 5], tmp[MAXN + 5], S[MAXN + 5], l = -1e6, r = 1e6;

inline bool check(double mid) {
	for (int i = 1; i <= n; ++i)
		tmp[i] = a[i] - mid, S[i] = S[i - 1] + tmp[i];
	
	double minv = INF, ret = -INF;
	
	for (int i = f; i <= n; ++i) {
		minv = min(minv, S[i - f]);
		ret = max(ret, S[i] - minv);
	}
	
	return ret >= 0;
}

int main() {
	scanf("%d%d", &n, &f);
	
	for (int i = 1; i <= n; ++i) 
		scanf("%lf", &a[i]);
		
	while (r - l > eps) {
		double mid = (l + r) / 2;
		
		if (check(mid)) 
			l = mid;
		else
			r = mid;
	}
	
	printf("%d\n", int(r * 1000));
	
	return 0;
}

BZOJ2002

\(LCT\)好题（大雾），可能因为我太懒了，不想打分块。

这道题关键在于\(LCT\)要求节点的深度，随便\(PushUp\)一下就好了。

#include <iostream>
#include <cstdio>
#define ls ch[0]
#define rs ch[1]
#define lson s[x].ls
#define rson s[x].rs
#define father s[x].fa
const int MAXN = 200000;
using namespace std;

inline int read() {
	int x = 0, w = 1; char c = ' ';
	while (c < '0' || c > '9') {c = getchar(); if (c == '-') w = -1;}
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
	return x * w;
}

struct Node {
	int ch[2], fa, rev, sz;
};

class LCT {
	public :
		void PushDown(int x) {
			if (s[x].rev) {
				s[lson].rev ^= 1, s[rson].rev ^= 1, s[x].rev ^= 1;
				swap(lson, rson);
			}
		}
		void PushUp(int x) {
			s[x].sz = s[lson].sz + s[rson].sz + 1;
		}
		bool IsRoot(int x);
		void rotate(int x); void Splay(int x);
		void access(int x); void MakeRoot(int x);
		void link(int x, int y); void cut(int x, int y);
		int query(int x, int y);
	private :
		Node s[MAXN + 5];
		int top, ST[MAXN + 5];
}T;

bool LCT::IsRoot(int x) {
	return s[father].ls != x && s[father].rs != x;
}

void LCT::rotate(int x) {
	int y = father, z = s[y].fa, l, r;
	if (s[y].ls == x) l = 0; else l = 1;
	r = l ^ 1;
	
	if (!IsRoot(y)) 
		if (s[z].ls == y) s[z].ls = x; else s[z].rs = x;
	father = z, s[y].fa = x, s[ s[x].ch[r] ].fa = y, s[y].ch[l] = s[x].ch[r], s[x].ch[r] = y;
	PushUp(y), PushUp(x);
}  

void LCT::Splay(int x) {
	top = 0, ST[++top] = x;
	for (int i = x; !IsRoot(i); i = s[i].fa) ST[++top] = s[i].fa;	
	for (int i = top; i; --i) PushDown(ST[i]);
	
	while (!IsRoot(x)) {
		int y = father, z = s[y].fa;
		if (!IsRoot(y))
			if (s[y].ls == x ^ s[z].ls == y) rotate(x); else rotate(y);
		rotate(x);
	}
}

void LCT::access(int x) {
	for (int last = 0; x; last = x, x = father)
		Splay(x), rson = last, PushUp(x);
}

void LCT::MakeRoot(int x) {
	access(x), Splay(x), s[x].rev ^= 1;
}

void LCT::link(int x, int y) {
	MakeRoot(x), father = y;
}

void LCT::cut(int x, int y) {
	MakeRoot(x), access(y), Splay(y);
	if (s[y].ls == x) s[y].ls = father = 0;
}

int LCT::query(int x, int y) {
	MakeRoot(x), access(y), Splay(y);
	return s[y].sz - 1;
}
 
int k[MAXN + 5];
 
int main() {
	int n = read();
	
	for (int i = 1; i <= n; ++i) {
 		k[i] = read(); int v = k[i] + i <= n ? k[i] + i : n + 1;
 		T.link(i, v);	
 	}
	
	int m = read();
	
	for (int i = 1; i <= m; ++i) {
		int opt = read(), u = read() + 1, v = u + k[u] <= n ? u + k[u] : n + 1, num;
		
		switch(opt) {
			case 1 : printf("%d\n", T.query(u, n + 1)); break;
			case 2 : {
				T.cut(u, v); num = read(), v = u + num <= n ? u + num : n + 1;
				T.link(u, v), k[u] = num;
				break;
			}
		}
	}
	
	return 0;
}