# 【BZOJ 1018】 [SHOI2008]堵塞的交通traffic

【题目链接】:http://www.lydsy.com/JudgeOnline/problem.php?id=1018

【题意】

【题解】

a[x][y]表示这个区间的左端点的上下方和右端点的上下方的连通性;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 110;
const int M = 1e5 + 100;

int n;
bool a[M][2];

struct abcd
{
bool a[2][2];
abcd(bool _ = false)
{
a[0][0] = a[1][1] = true;
a[0][1] = a[1][0] = _;
}
bool* operator [] (int x)
{
return a[x];
}
friend abcd merge(bool sta[2], abcd x, abcd y)
{
abcd re;
re[0][0] = (x[0][0] & sta[0] & y[0][0]) | (x[0][1] & sta[1] & y[1][0]);
re[1][1] = (x[1][1] & sta[1] & y[1][1]) | (x[1][0] & sta[0] & y[0][1]);
re[0][1] = (x[0][0] & sta[0] & y[0][1]) | (x[0][1] & sta[1] & y[1][1]);
re[1][0] = (x[1][0] & sta[0] & y[0][0]) | (x[1][1] & sta[1] & y[1][0]);
return re;
}
};

struct segtree
{
segtree *ls, *rs;
abcd status;
segtree() :ls(0x0), rs(0x0) {}
#define push_up(); status = merge(a[mid],ls->status,rs->status);
void Build_Tree(int x, int y)
{
if (x == y) return;
int mid = (x + y) >> 1;
ls = new segtree, rs = new segtree;
ls->Build_Tree(x, mid);
rs->Build_Tree(mid + 1, y);
push_up();
}
void refresh(int x, int y, int pos)
{
int mid = (x + y) >> 1;
if (pos == mid)
{
push_up();
return;
}
if (pos < mid)
ls->refresh(x, mid, pos);
else
rs->refresh(mid + 1, y, pos);
push_up();
}
void modify(int x, int y, int pos, int flag)
{
int mid = (x + y) >> 1;
if (x == y)
{
new(&status) abcd(flag);
return;
}
if (pos <= mid)
ls->modify(x, mid, pos, flag);
else
rs->modify(mid + 1, y, pos, flag);
push_up();
}
void _get_left(int x, int y, int &pos, abcd &sta, int flag)
{
int mid = (x + y) >> 1;
if (x == y) return;
abcd temp = merge(a[y], rs->status, sta);
if (temp[0][flag] || temp[1][flag])
pos = mid + 1, sta = temp, ls->_get_left(x, mid, pos, sta, flag);
else
rs->_get_left(mid + 1, y, pos, sta, flag);
}
void get_left(int x, int y, int &pos, abcd &sta, int flag)
{
if (x == y) return;
int mid = (x + y) >> 1;
if (pos <= mid)
ls->get_left(x, mid, pos, sta, flag);
else
{
//pos > mid
rs->get_left(mid + 1, y, pos, sta, flag);
if (pos != mid + 1) return;
//pos==mid+1;
abcd temp = merge(a[mid], ls->status, sta);
if (temp[0][flag] || temp[1][flag])
pos = x, sta = temp;
else
ls->_get_left(x, mid, pos, sta, flag);
}
}
void _get_right(int x, int y, int &pos, abcd &sta, int flag)
{
int mid = (x + y) >> 1;
if (x == y) return;
abcd temp = merge(a[x - 1], sta, ls->status);
if (temp[flag][0] || temp[flag][1])
pos = mid, sta = temp, rs->_get_right(mid + 1, y, pos, sta, flag);
else
ls->_get_right(x, mid, pos, sta, flag);
}
void get_right(int x, int y, int &pos, abcd &sta, int flag)
{
if (x == y) return;
int mid = (x + y) >> 1;
if (mid < pos)
rs->get_right(mid + 1, y, pos, sta, flag);
else
{
//pos <= mid
ls->get_right(x, mid, pos, sta, flag);
if (pos != mid) return;
//pos==mid
abcd temp = merge(a[mid], sta, rs->status);
if (temp[flag][0] || temp[flag][1])
pos = y, sta = temp;
else
rs->_get_right(mid + 1, y, pos, sta, flag);
}
}
abcd get_ans(int x, int y, int l, int r)
{
if (x == l && y == r)
return status;
int mid = (x + y) >> 1;
if (mid < l)
return rs->get_ans(mid + 1, y, l, r);
if (r <= mid)
return ls->get_ans(x, mid, l, r);
return merge(a[mid], ls->get_ans(x, mid, l, mid), rs->get_ans(mid + 1, y, mid + 1, r));
}
} *tree = new segtree;

void modify(int x1, int y1, int x2, int y2, bool flag)
{
if (x1 == x2)
{
if (y1 > y2) swap(y1, y2);
a[y1][x1 - 1] = flag;
tree->refresh(1, n, y1);
return;
}
//y1 == y2
tree->modify(1, n, y1, flag);
}

void query(int x1, int y1, int x2, int y2)
{
if (y1 > y2)
{
swap(x1, x2);
swap(y1, y2);
}
abcd temp(false);
tree->get_left(1, n, y1, temp, x1 - 1);
x1 = temp[0][x1 - 1] ? 1 : 2;

abcd temp1(false);
tree->get_right(1, n, y2, temp1, x2 - 1);
x2 = temp1[x2 - 1][0] ? 1 : 2;

abcd temp2 = tree->get_ans(1, n, y1, y2);
puts(temp2[x1 - 1][x2 - 1] ? "Y" : "N");
}

int main()
{
//freopen("F:\\rush.txt", "r", stdin);
rei(n);
tree->Build_Tree(1, n);
int x1, y1, x2, y2;
char p[10];
while (1)
{
scanf("%s", p);
rei(x1), rei(y1), rei(x2), rei(y2);
if (p[0] == 'C')
modify(x1, y1, x2, y2, false);
if (p[0] == 'O')
modify(x1, y1, x2, y2, true);
if (p[0] == 'A')
query(x1, y1, x2, y2);
if (p[0] == 'E')
break;
}
return 0;
}
posted @ 2017-10-04 18:45  AWCXV  阅读(65)  评论(0编辑  收藏