UVALive 3959 Rectangular Polygons （排序贪心）

Rectangular Polygons

题目链接：

http://acm.hust.edu.cn/vjudge/contest/129733#problem/G

Description

In this problem, we will help the Faculty of Civil Engineering. They need a software to analyze ground plans of buildings. Specifically, your task is to detect outlines of a building when all of its corners are given.
You may assume that each building is a rectangular polygon with each of its sides being parallel either with X or Y axis. Therefore, each of its vertex angles is exactly either 90 or 270 degrees.

Input

The input contains several buildings. The description of each building starts with a single positive integer N , the number of corners (polygon vertices), 1

Output

For each building, output one line containing N characters without any whitespace between them. The characters should be uppercase letters that specify directions of individual walls (sides) when the building outline is followed. " N" stands for North (the positive direction of the Y axis), `` E" for East (the positive direction of the X axis), "W" for West, and " S" for South. The "walk" should start in the vertex that has been given first in the input and always proceed in the clockwise direction.

Sample Input

4
0 0
2 2
0 2
2 0
6
1 1
2 2
0 1
1 0
0 2
2 0
0

Sample Output

NESW
WNESWN

Source

2016-HUST-线下组队赛-2


##题意：

给出平面上N个点，构造一个每条边都平行于轴的多边形.
从起点#1开始顺时针输出行进方向.


##题解：

观察结果图形：所有边一定平行轴. 那么对于y坐标相同的若干点#1~#m，有若干推论：（均可以反证）
首先，m的一定是偶数，否者多出来的点不能正常连边.
然后，这m个点一定有 m/2 条边，否则剩下的边不能正常连边.
最后，这m个点的连边方式一定是：(1,2),(3,4)...(m-1,m)，否则会出现交叉或多余点.
对于x坐标相同的点也是如此.
对所有点按x为第一优先级排序后，可以将所有竖边连起来；同理得到所有横边.

现在已经得到关于最后结果的图模型，问题是怎么按要求输出：
直接按从#1开始或许有点麻烦，这里可以从最左上角的点开始往右走，这样保证了输出的轨迹一定是顺时针.
当遍历到#1时才开始输出，计数n次输出即结束. 所以最多遍历2次.

这题瞎搞了两小时，写了160+行结果TLE... 赛时想得太复杂，各种xjb删点删边瞎搞一通.
今天真是打得太挫了，2个小时写两题，后面的时间都在搞一个题. 两个人搞还是有点蛋疼，有几道比较简单的题都没开.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 1010
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

struct Point {

int id;

int x, y;

}p[maxn];

int g[maxn][4];

bool cmp1(Point a, Point b) {

if(a.y == b.y) return a.x < b.x;

return a.y < b.y;

}

bool cmp2(Point a, Point b) {

if(a.x == b.x) return a.y < b.y;

return a.x < b.x;

}

int main(int argc, char const *argv[])

{

//IN;

while(scanf("%d", &n) != EOF && n)
{
    memset(g, 0, sizeof(g));
    int cur = 0, cx = inf, cy = -inf; //top-left most
    for(int i=1; i<=n; i++) {
        scanf("%d %d", &p[i].x,&p[i].y);
        p[i].id = i;
        if(p[i].y > cy) cx = p[i].x, cy = p[i].y, cur = i;
        if(p[i].y == cy && p[i].x < cx) cx = p[i].x, cur = i;
    }

    sort(p+1, p+1+n, cmp1);
    for(int i=1; i<=n; i+=2) {
        g[p[i].id][1] = p[i+1].id;
        g[p[i+1].id][3] = p[i].id;
    }

    sort(p+1, p+1+n, cmp2);
    for(int i=1; i<=n; i+=2) {
        g[p[i].id][0] = p[i+1].id;
        g[p[i+1].id][2] = p[i].id;
    }

    int order = 0;
    bool flag = 0;
    int cnt = 0;
    while(1) {
        if(cur == 1) flag = 1;

        char ans = 0;
        int tmp = cur;
        if(!order) {
            if(g[tmp][1]) cur = g[tmp][1], ans = 'E';
            if(g[tmp][3]) cur = g[tmp][3], ans = 'W';
        } else {
            if(g[tmp][0]) cur = g[tmp][0], ans = 'N';
            if(g[tmp][2]) cur = g[tmp][2], ans = 'S';
        }
        order = !order;

        if(flag) putchar(ans), cnt++;
        if(cnt >= n) break;
    }
    putchar('\n');
}

return 0;

}