POJ3109-Inner Vertices

Inner Vertices

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 2583 Accepted: 739Case Time Limit: 2000MS

Description

There is an infinite square grid. Some vertices of the grid are black and other vertices are white.

A vertex V is called inner if it is both vertical-inner and horizontal-inner. A vertex V is called horizontal-inner if there are two such black vertices in the same row that V is located between them. A vertex V is called vertical-inner if there are two such black vertices in the same column that V is located between them.

On each step all white inner vertices became black while the other vertices preserve their colors. The process stops when all the inner vertices are black.

Write a program that calculates a number of black vertices after the process stops.

Input

The first line of the input file contains one integer number n (0 ≤ n ≤ 100 000) — number of black vertices at the beginning.

The following n lines contain two integer numbers each — the coordinates of different black vertices. The coordinates do not exceed 10⁹ by their absolute values.

Output

Output the number of black vertices when the process stops. If the process does not stop, output -1.

Sample Input

40 22 0-2 00 -2

Sample Output

5

Hint

Source

Northeastern Europe 2005, Northern Subregion

题意：一个无限大的棋盘上有无数个点，这些点有黑有白，如果有一个点，上下左右各有点，则将这个点涂成黑色。最后计算这个棋盘上黑点的数量

解题思路：先将坐标离散化，然后判断出对于每个x轴的最高的y。对所有点可以按y轴从小到大进行排序，y轴相同的按x轴从小到大进行排序。每扫到一个点，若这个点不是这个x轴的最高的y，则在线段树上将这个位置置为一，否则的话置为0。扫的时候若前后两个点的y坐标一样，则查询一下在这两个x轴之间区间值为多少，则答案就加多少

#include <iostream>    #include <cstdio>    #include <string>    #include <cstring>    #include <algorithm>    #include <cmath>    #include <vector>    #include <map>    #include <set>    #include <queue>    #include <stack>    #include <functional>    #include <climits>    using namespace std;#define LL long long    const int INF = 0x3f3f3f3f;int n, x[100009], y[100009], sum[100009 << 2], f[100009];struct point{int x, y;bool operator<(const point &b)const{if (y != b.y) return y < b.y;else return x < b.x;}}p[100009];void update(int k, int l, int r, int p, int val){if (l == r) { sum[k] = val; return; }int mid = (l + r) >> 1;if (mid >= p) update(k << 1, l, mid, p, val);else update(k << 1 | 1, mid + 1, r, p, val);sum[k] = sum[k << 1] + sum[k << 1 | 1];}int query(int k, int l, int r, int ll, int rr){if (l >= ll&&r <= rr) return sum[k];int mid = (l + r) >> 1, ans = 0;if (ll <= mid) ans += query(k << 1, l, mid, ll, rr);if (rr > mid) ans += query(k << 1 | 1, mid + 1, r, ll, rr);return ans;}int main(){while (~scanf("%d", &n)){memset(f, 0, sizeof f);int cnt1 = 1, cnt2 = 1;for (int i = 1; i <= n; i++){scanf("%d%d", &p[i].x, &p[i].y);x[cnt1++] = p[i].x, y[cnt2++] = p[i].y;}sort(p + 1, p + 1 + n);sort(x + 1, x + cnt1);sort(y + 1, y + cnt2);cnt1 = unique(x + 1, x + cnt1) - x;cnt2 = unique(y + 1, y + cnt2) - y;memset(sum, 0, sizeof sum);for (int i = n; i >= 1; i--){int xx = lower_bound(x + 1, x + cnt1, p[i].x) - x;int yy = lower_bound(y + 1, y + cnt2, p[i].y) - y;if (f[xx]) continue;f[xx] = yy;}int ans = 0;for (int i = 1; i <= n; i++){int xx = lower_bound(x + 1, x + cnt1, p[i].x) - x;int yy = lower_bound(y + 1, y + cnt2, p[i].y) - y;if (f[xx] > yy) update(1, 1, cnt1, xx, 1);else update(1, 1, cnt1, xx, 0);if (i == 1 || p[i].y != p[i - 1].y) continue;int l = lower_bound(x + 1, x + cnt1, p[i - 1].x) - x + 1;int r = xx - 1;if (l > r) continue;ans += query(1, 1, cnt1, l, r);}printf("%d\n", ans + n);}return 0;}