bzoj1050

本文深入探讨了强行LCT(链剖线树)算法的实现细节,包括节点维护、路径反转、旋转操作等核心概念,并通过一个具体示例展示了如何使用该算法解决特定问题。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

强行lct    


貌似bzoj数据不对?  有自环 (逃


#include <cstdio>
#include <algorithm>

using namespace std;

int ch[5510][2],fa[5510],rv[5510],x[5510],nu[5510],mi[5510],mn[5510];
int st[5510],tp;

int n,m,ST,EN;

int head[510];

struct edge {
	int u,v,w;
	bool operator < (const edge &b)const {
		return w < b.w;
	}
}e[5010];

int read_int () {
	char c = getchar();
	int re = 0;
	for(;c > '9' || c < '0';c = getchar());
	for(;c >= '0' && c <= '9';c = getchar())
		re = re * 10 + c - '0';
	return re;
}

int find (int u) {
	if(head[u] == u)
		return u;
	return head[u] = find(head[u]);
}

void pu (int u) {
	mi[u] = x[u];
	mn[u] = nu[u];
	if(mi[u] > mi[ch[u][0]]) {
		mi[u] = mi[ch[u][0]];
		mn[u] = mn[ch[u][0]];
	}
	if(mi[u] > mi[ch[u][1]]) {
		mi[u] = mi[ch[u][1]];
		mn[u] = mn[ch[u][1]];
	}
}

void reverse (int u) {
	int t = ch[u][0];
	ch[u][0] = ch[u][1];
	ch[u][1] = t;
	rv[u] ^= 1;
}

void pd (int u) {
	if(!rv[u])
		return;
	rv[u] = 0;
	reverse(ch[u][0]);
	reverse(ch[u][1]);
}

void rot (int u,int d) {
	ch[fa[u]][d ^ 1] = ch[u][d];
	if(ch[u][d])
		fa[ch[u][d]] = fa[u];
	int t = fa[fa[u]];
	fa[fa[u]] = u;
	ch[u][d] = fa[u];
	fa[u] = t;
	
	pu(ch[u][d]);
	pu(u);
	
	if(!t)
		return;
	if(ch[t][0] == ch[u][d])
		ch[t][0] = u;
	if(ch[t][1] == ch[u][d])
		ch[t][1] = u;
}

void splay (int u) {
	st[tp = 1] = u;
	int t = u;
	while(ch[fa[t]][0] == t || ch[fa[t]][1] == t) {
		t = fa[t];
		st[++tp] = t;
	}
	while(tp)
		pd(st[tp--]);
		
	while(ch[fa[u]][0] == u || ch[fa[u]][1] == u) {
		if(ch[fa[fa[u]]][0] != fa[u] && ch[fa[fa[u]]][1] != fa[u]) {
			if(ch[fa[u]][0] == u)
				rot(u,1);
			else rot(u,0);
		}
		else {
			int d = ch[fa[fa[u]]][0] == fa[u];
			if(ch[fa[u]][d] == u) {
				rot(u,d ^ 1);
				rot(u,d);
			}
			else {
				rot(u,d);
				rot(u,d);
			}
		}
	}
}

void access (int u) {
	int v = 0;
	while(u) {
		splay(u);
		ch[u][1] = v;
		pu(u);
		v = u;
		u = fa[u];
	}
}

void mtr (int u) {
	access(u);
	splay(u);
	reverse(u);
}

void link (int u,int v) {
	mtr(u);
	mtr(v);
	fa[v] = u;
}

void cut (int u,int v) {
	mtr(u);
	access(u);
	splay(v);
	fa[v] = 0;
}

int gcd (int a,int b) {
	if(a % b)
		return gcd(b,a % b);
	return b;
}

int main () {
	
	n = read_int();
	m = read_int();
	
	for(int i = 0;i <= n;++i) {
		head[i] = i;
		mi[i] = x[i] = 0x3f3f3f3f;
	}
	
	for(int i = 1;i <= m;++i) {
		e[i].u = read_int();
		e[i].v = read_int();
		e[i].w = read_int();
	}
	
	ST = read_int();
	EN = read_int();
	
	sort(e + 1,e + 1 + m);
	
	int a1 = 0x3f3f3f3f,a2 = 1;
	
	
	for(int i = 1;i <= m;++i) {
//	this part is different 
		if(e[i].u != e[i].v) {
			if(find(e[i].u) == find(e[i].v)) {
				mtr(e[i].u);
				access(e[i].v);
				splay(e[i].u);
				int t = mn[e[i].u];
				cut(t + n,e[t].u);
				cut(t + n,e[t].v);
			}
			else head[find(e[i].u)] = find(e[i].v);
			x[i + n] = mi[i + n] = e[i].w;
			mn[i + n] = nu[i + n] = i;
			link(e[i].u,i + n);
			link(e[i].v,i + n);
		}
		
		if(find(ST) == find(EN)) {
			mtr(ST);
			access(EN);
			splay(ST);
			if((double)e[i].w / (double)mi[ST] < (double)a1 / (double)a2) {
				a1 = e[i].w / gcd(e[i].w,mi[ST]);
				a2 = mi[ST] / gcd(e[i].w,mi[ST]);
			}
		}
	}
	if(a1 == 0x3f3f3f3f)
		printf("IMPOSSIBLE\n");
	else {
		if(a1 / a2 * a2 == a1)
			printf("%d\n",a1 / a2);
		else printf("%d/%d\n",a1,a2);
	}
	return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值