【BZOJ1018】[SHOI2008]堵塞的交通

【BZOJ1018】[SHOI2008]堵塞的交通

题面

bzoj

洛谷

题解

菊队讲要用线段树维护连通性，但是好像没人写

解法一

将所有的加边删边离线，然后以最近删除时间为边权，$LCT$维护最大生成树即可

代码

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm>
#include <map> 
using namespace std; 
inline int gi() {
	register int data = 0, w = 1;
	register char ch = 0;
	while (!isdigit(ch) && ch != '-') ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar();
	while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
	return w * data; 
}
const int MAX_N = 100005;
const int INF = 1e9; 
struct Node { int fa, ch[2], v, d; bool rev; } t[MAX_N << 2];
int stk[MAX_N << 2], top; 
void pushup(int x) { 
	t[x].d = x; 
	if (t[t[x].d].v > t[t[t[x].ch[0]].d].v) t[x].d = t[t[x].ch[0]].d; 
	if (t[t[x].d].v > t[t[t[x].ch[1]].d].v) t[x].d = t[t[x].ch[1]].d; 
}
void pushrev(int x) {
	t[x].rev ^= 1;
	swap(t[x].ch[0], t[x].ch[1]); 
} 
void pushdown(int x) {
	if (!t[x].rev) return ; 
	if (t[x].ch[0]) pushrev(t[x].ch[0]);
	if (t[x].ch[1]) pushrev(t[x].ch[1]);
	t[x].rev = 0; 
} 
int get(int x) { return t[t[x].fa].ch[1] == x; }
bool isroot(int x) { return (t[t[x].fa].ch[1] != x) && (t[t[x].fa].ch[0] != x); } 
void rotate(int x) {
	int k = get(x), y = t[x].fa, z = t[y].fa;
	if (!isroot(y)) t[z].ch[get(y)] = x;
	t[x].fa = z; 
	t[t[x].ch[k ^ 1]].fa = y, t[y].ch[k] = t[x].ch[k ^ 1];
	t[x].ch[k ^ 1] = y, t[y].fa = x;
	pushup(y), pushup(x); 
}
void splay(int x) { 
	stk[top = 1] = x;
	for (int i = x; !isroot(i); i = t[i].fa) stk[++top] = t[i].fa; 
	for (int i = top; i; i--) pushdown(stk[i]);
	while (!isroot(x)) {
		int y = t[x].fa;
		if (!isroot(y)) (get(x) ^ get(y)) ? rotate(x) : rotate(y);
		rotate(x); 
	} 
}
void access(int x) { for (int y = 0; x; y = x, x = t[x].fa) splay(x), t[x].ch[1] = y, pushup(x); }
int findroot(int x) { access(x); splay(x); while (t[x].ch[0]) pushdown(x), x = t[x].ch[0]; return x; }
void makeroot(int x) { access(x); splay(x); pushrev(x); } 
void split(int x, int y) { makeroot(x); access(y); splay(y); } 
void link(int x, int y) { makeroot(x); t[x].fa = y; } 
void cut(int x, int y) { split(x, y); t[y].ch[0] = t[x].fa = 0, pushup(y); }
int N, C, tot; 
struct Opt { int t, op, x1, x2, del; } a[MAX_N << 1]; 
map<pair<int, int>, int> mp;

int main () {
	C = gi(); 
    for ( ; ; )  {
		char ch[10]; scanf("%s", ch);
		if (ch[0] == 'E') break; 
		int op, r1 = gi() - 1, c1 = gi(), r2 = gi() - 1, c2 = gi(), x1 = r1 * C + c1, x2 = r2 * C + c2;
		if (x1 > x2) swap(x1, x2); 
		if (ch[0] == 'O') op = 0;
		if (ch[0] == 'C') op = 1;
		if (ch[0] == 'A') op = 2;
		++N; 
		if (op == 0) mp[pair<int, int>(x1, x2)] = N, a[N].del = INF; 
		if (op == 1) a[mp[pair<int, int>(x1, x2)]].del = N; 
		a[N].t = N, a[N].op = op, a[N].x1 = x1, a[N].x2 = x2; 
	} 
	tot = 2 * C;
	for (int i = 0; i <= tot; i++) t[i].v = INF; 
	for (int i = 1; i <= N; i++) {
		int x1 = a[i].x1, x2 = a[i].x2, del = a[i].del; 
		if (a[i].op == 0) { 
			if (findroot(x1) == findroot(x2)) {
				split(x1, x2); int d = t[x2].d; 
				if (t[d].v >= del) continue;
				else cut(x1, d), cut(x2, d);
				t[++tot].v = del;
				link(x1, tot), link(x2, tot); 
			}
			else t[++tot].v = del, link(x1, tot), link(x2, tot); 
		}
	    if (a[i].op == 1)
			if (findroot(x1) == findroot(x2)) { 
			    split(x1, x2); 
				int d = t[x2].d; 
				if (t[d].v > a[i].t) continue; 
				cut(x1, d), cut(x2, d); 
			}
		if (a[i].op == 2) {
			if (findroot(x1) == findroot(x2)) puts("Y");
			else puts("N"); 
		} 
	} 
	return 0; 
} 

解法二

线段树分治板子题，然而我并不会

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