UVA 11106 - Rectilinear Polygon（几何+贪心）

Problem B: Rectilinear polygon

Given is n  points with integer coordinates in the plane. Is it is possible to construct a simple, that is non-intersecting, rectilinear polygon with the given points as vertices? In a rectilinear polygon there are at least 4 vertices and every edge is either horizontal or vertical; each vertex is an endpoint of exactly one horizontal edge and one vertical edge. There are no holes in a polygon.

The first line of input is an integer giving the number of cases that follow. The input of each case starts with an integer 4 ≤ n≤ 100000 giving the number of points for this test case. It is followed by n pairs of integers specifying the x and ycoordinates of the points for this case.

The output should contain one line for each case on input. Each line should contain one integer number giving the length of the rectilinear polygon passing throught the given points when it exists; otherwise, it should contain -1.

Sample input

1
8
1 2
1 0
2 1
2 2
3 2
3 1
4 0
4 2

Output for sample input

12

题意：给定n个点，判断这n点能不能组成一个多边形，并且点都在直角上，如果可以输出周长，不行输出-1.

思路：对于横坐标为x的所有点，如果是奇数个点，那么肯定不行，因为同一横坐标x的点，要能折出去再折回来，必然为偶数个，按x优先，y第二优先排序后，组成的边必然为，0,1组，2,3组，3,4组...对于纵坐标也是同理，这样就能求出周长了。然后在去判断有没有自交，判断方式为，先把所有以x求出来的线段保存下来，然后在求y的时候，去判断有没有相交的边（这步用了map，不过跑起来时间还是很短）。最后还要判断组合出来的多边形是否只有一个。判断方式是，之前求线段的时候，把每个点水平和垂直连的点都求出来，随便选一点进行dfs，判断最后形成环的时候，点的个数等不等于n。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <map>
#include <algorithm>
using namespace std;

const int N = 100005;
typedef pair<int, int> pi;
map<int, pi> vis;
int t, n, en, vi[N], num;
struct Point {
	int x, y, id, v, h;
} p[N];

struct Edge {
	int x, y1, y2;
} e[N];

bool cmp1(Point a, Point b) {
	if (a.x != b.x)
		return a.x < b.x;
	return a.y < b.y;
}

bool cmp2(Point a, Point b) {
	if (a.y != b.y)
		return a.y < b.y;
	return a.x < b.x;
}

bool cmp3(Point a, Point b) {
	return a.id < b.id;
}

void init() {
	en = 0; vis.clear();
	memset(vi, 0, sizeof(vi));
	num = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &p[i].x, &p[i].y);
		p[i].id = i;
	}
}

void dfs(int id, int bo) {
	if (vi[id]) return;
	vi[id] = 1; num++;
	if (bo) dfs(p[id].v, 0);
	else dfs(p[id].h, 1);
}

int solve() {
	if (n % 2) return -1;
	int ans = 0;
	sort(p, p + n, cmp1); int i, save = -1;
	for (i = 0; i < n; i += 2) {
		if (p[i].x != p[i + 1].x) return -1;
		if (save != p[i].x) {
			vis[p[i].x].first = en;
			if (i != 0) vis[p[i - 1].x].second = en;
			save = p[i].x;
		}
		e[en].x = p[i].x; e[en].y1 = p[i + 1].y; e[en++].y2 = p[i].y;
		p[i].h = p[i + 1].id;
		p[i + 1].h = p[i].id;
		ans += (p[i + 1].y - p[i].y);
	}
	vis[p[n - 1].x].second = en;
	sort(p, p + n, cmp2);
	for (i = 0; i < n; i += 2) {
		if (p[i].y != p[i + 1].y) return -1;
		for (int j = vis[p[i].x].second; j < vis[p[i + 1].x].first; j++) {
			if (e[j].y1 >= p[i].y && e[j].y2 <= p[i].y) return -1;
		}
		p[i].v = p[i + 1].id;
		p[i + 1].v = p[i].id;
		ans += (p[i + 1].x - p[i].x);
	}
	sort(p, p + n, cmp3);
	dfs(0, 0);
	if (num != n) return -1;
	return ans;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		init();
		printf("%d\n", solve());
	}
	return 0;
}