bzoj 1018 堵塞的交通traffic

题外话

做了这个线段树的题我整个人都不好了，头一次做这种用线段树维护连通性的题，简直烦的要死= =

Description

给你一个2*n的格子，开始全不联通，相邻两点可以连边，有3种操作

$1:将(x1,y1),(x2,y2)变为连通$

$2:将(x1,y1),(x2,y2)变为不连通$

$3:询问(x1,y1),(x2,y2)是否联通$

Soluion

用线段树维护6个信息，正常的4条边以及两条对角线是否联通，修改时用线段树维护

最蛋疼的是查询，比如$[l,r]区间$，不一定连通性只是在这个区间内的

比如虽然$[l,r]$，虽然不连通，但是可以在$[1,l]$区间能从l上走到下，则l这个位置上下就是联通的，右侧同理，细心处理就可以了

Code

#include <bits/stdc++.h>//good
#define ls (rt << 1)
#define rs (rt << 1 | 1)
using namespace std;
const int N = 100010;
typedef long long LL;
int n;
char s[10];
struct Node {
    int l, r, h0, h1, s0, s1, x0, x1;
}a[N << 2];
int g[2][N][4];//U D L R
inline int read(int &t) {
    int f = 1;char c;
    while (c = getchar(), c < ‘0‘ || c > ‘9‘) if (c == ‘-‘) f = -1;
    t = c - ‘0‘;
    while (c = getchar(), c >= ‘0‘ && c <= ‘9‘) t = t * 10 + c - ‘0‘;
    t *= f;
}
Node merge(Node a, Node b) {
    Node c;
    int f0 = g[0][a.r][3], f1 = g[1][a.r][3];
    c.h0 = (a.h0 && f0 && b.h0) | (a.x0 && f1 && b.x1);
    c.h1 = (a.h1 && f1 && b.h1) | (a.x1 && f0 && b.x0);
    c.s0 = a.s0 | (a.h0 && f0 && b.s0 && f1 && a.h1);
    c.s1 = b.s1 | (b.h0 && f0 && a.s1 && f1 && b.h1);
    c.x0 = (a.h0 && f0 && b.x0) | (a.x0 && f1 && b.h1);
    c.x1 = (a.h1 && f1 && b.x1) | (a.x1 && f0 && b.h0);
    c.l = a.l, c.r = b.r;
    return c;
}
void modify(int x1, int y1, int x2, int y2, int x) {
    if (x1 == x2)   g[x1][y1][3] = g[x1][y2][2] = x;
    else g[0][y1][1] = g[1][y1][0] = x;
}
void build(int rt, int l, int r) {
    a[rt].l = l, a[rt].r = r;
    if (l == r) {
        a[rt].h0 = a[rt].h1 = 1;
        return;
    }
    int mid = l + r >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
}
void change(int rt, int l, int r, int x1, int y1, int x2, int y2) {
    int mid = l + r >> 1;
    if (x1 == x2 && mid == y1) {
        a[rt] = merge(a[ls], a[rs]);
        return;
    }
    if (l == r && y1 == y2) {
        a[rt].x0 = a[rt].x1 = 0;
        a[rt].s0 = a[rt].s1 = g[0][y1][1];
        if (a[rt].s0)   a[rt].x0 = a[rt].x1 = 1;
        return ;
    }
    if (y1 <= mid)  change(ls, l, mid, x1, y1, x2, y2);
    else change(rs, mid + 1, r, x1, y1, x2, y2);
    a[rt] = merge(a[ls], a[rs]);
}
Node find(int rt, int l, int r) {
    if (l <= a[rt].l && a[rt].r <= r)   return a[rt];
    int mid = a[rt].l + a[rt].r >> 1;
    if (r <= mid)   return find(ls, l, r);
    if (l > mid)    return find(rs, l, r);
    else return merge(find(ls, l, r), find(rs, l, r));
}
bool ask(int x1, int y1, int x2, int y2) {
    Node a = find(1, 1, y1), b = find(1, y1, y2), c = find(1, y2, n);
    b.s0 |= a.s1, b.s1 |= c.s0;
    if (x1 == x2) {
        if (x1 == 0)    return b.h0 | (b.s0 && b.h1 && b.s1) | (b.s0 && b.x1) | (b.x0 && b.s1);
        else return b.h1 | (b.s0 && b.h0 && b.s1) | (b.s0 && b.x0) | (b.x1 && b.s1);
    }
    else {
        if (x1 < x2)    return b.x0 | (b.h0 && b.s1) | (b.s0 && b.h1);
        else return b.x1 | (b.h1 && b.s1) | (b.s0 && b.h0);
    }   
}
int main(){
    read(n);
    build(1, 1, n);
    while (scanf("%s", s) && s[0] != ‘E‘) {
        int x1, y1, x2, y2;
        read(x1), read(y1), read(x2), read(y2);
        --x1, --x2;
        if (s[0] == ‘C‘) {
            if (y1 > y2)    swap(y1, y2);
            modify(x1, y1, x2, y2, 0);
            change(1, 1, n, x1, y1, x2, y2);
        }
        else if (s[0] == ‘O‘) {
            if (y1 > y2)    swap(y1, y2);
            modify(x1, y1, x2, y2, 1);
            change(1, 1, n, x1, y1, x2, y2);
        }
        else {
            if (y1 > y2)    swap(x1, x2), swap(y1, y2);
            if (ask(x1, y1, x2, y2))    puts("Y");
            else puts("N");
        }
    }
    return 0;
}