标签：efi traffic query can 方式 target key break ns2
题意：有一个n * 2的网格图，有3种操作：给两个相邻的点加上一条边，断开相邻的两个点连着的边，询问两个点的连通性。
思路：直接看博客就行了，https://blog.csdn.net/roll_keyboard/article/details/81185535，在纸上画一画来确定4个顶点之间的更新关系。有一个需要注意的细节是有可能通过绕远路的方式可以到达，所以需要考虑前面和后面的部分对询问区间的影响。
代码：
#include <bits/stdc++.h>
#define ls (o << 1)
#define rs (o << 1 | 1)
using namespace std;
const int maxn = 100010;
struct node {
bool l, r, u, d, x1, x2, U, D;
};
node tr[maxn * 4];
node init() {
node ans;
ans.l = ans.r = ans.D = ans.U = ans.x1 = ans.x2 = 0;
ans.u = ans.d = 1;
return ans;
}
node merge(node a1, node a2) {
node ans;
ans.l = (a1.l | (a1.u & a1.U & a2.l & a1.D & a1.d));
ans.r = (a2.r | (a2.u & a1.U & a1.r & a1.D & a1.d));
ans.u = ((a1.u & a1.U & a2.u) | (a1.x1 & a1.D & a2.x2));
ans.d = ((a1.d & a1.D & a2.d) | (a1.x2 & a1.U & a2.x1));
ans.x1 = ((a1.u & a1.U & a2.x1) | (a1.x1 & a1.D & a2.d));
ans.x2 = ((a1.x2 & a1.U & a2.u) | (a1.d & a1.D & a2.x2));
ans.U = a2.U;
ans.D = a2.D;
return ans;
}
void build(int o, int l, int r) {
if(l == r) {
tr[o] = init();
return;
}
int mid = (l + r) >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
tr[o] = merge(tr[ls], tr[rs]);
}
void update(int o, int l, int r, int p, int type) {
if(l == r) {
if(type >> 2) {
tr[o].l = tr[o].r = tr[o].x1 = tr[o].x2 = (type & 1);
return;
} else if (type >> 1) {
tr[o].U = (type & 1);
} else {
tr[o].D = (type & 1);
}
return;
}
int mid = (l + r) >> 1;
if(p <= mid) update(ls, l, mid, p, type);
else update(rs, mid + 1, r, p, type);
tr[o] = merge(tr[ls], tr[rs]);
}
node query(int o, int l, int r, int ql, int qr) {
if(l >= ql && r <= qr) {
return tr[o];
}
int mid = (l + r) >> 1;
node ans = init();
ans.U = ans.D = 1;
if(ql <= mid) ans = merge(ans, query(ls, l, mid, ql, qr));
if(qr > mid) ans = merge(ans, query(rs, mid + 1, r, ql, qr));
return ans;
}
void out(bool flag) {
if(flag) printf("Y\n");
else printf("N\n");
}
char s[10];
int main() {
int n;
scanf("%d", &n);
int x1, y1, x2, y2;
build(1, 1, n);
while(1) {
scanf("%s", s + 1);
if(s[1] == ‘E‘) break;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if(y1 > y2) {
swap(x1, x2);
swap(y1, y2);
}
if(s[1] == ‘A‘) {
node ans1, ans2, ans3;
if(y1 > 1) ans1 = query(1, 1, n, 1, y1 - 1);
ans2 = query(1, 1, n, y1, y2);
if(y2 < n) ans3 = query(1, 1, n, y2 + 1, n);
if(y1 > 1 && (ans1.r && ans1.U && ans1.D)) {
ans2.l |= 1;
ans2.x1 |= (ans2.l & ans2.d);
ans2.x2 |= (ans2.l & ans2.u);
ans2.d |= (ans2.x1 & ans2.l);
ans2.u |= (ans2.l & ans2.x2);
}
if(y2 < n && (ans2.U && ans3.l && ans2.D)) {
ans2.r |= 1;
ans2.x1 |= (ans2.u & ans2.r);
ans2.x2 |= (ans2.d & ans2.r);
ans2.d |= (ans2.x2 & ans2.r);
ans2.u |= (ans2.x1 & ans2.r);
}
if(x1 == 1 && x2 == 1) {
out(ans2.u);
} else if(x1 == 1 && x2 == 2) {
out(ans2.x1);
} else if(x1 == 2 && x2 == 1) {
out(ans2.x2);
} else {
out(ans2.d);
}
} else {
int type = 0;
if(y1 == y2) {
type |= (1 << 2);
if(s[1] == ‘O‘) type |= 1;
} else {
if(x1 == 1) type |= (1 << 1);
if(s[1] == ‘O‘) type |= 1;
}
update(1, 1, n, y1, type);
}
}
}
标签：efi traffic query can 方式 target key break ns2
原文地址：https://www.cnblogs.com/pkgunboat/p/11445883.html
