UVA 11106

n

-1.

181 21 02 12 23 23 14 04 2

12

題意：給定n個點，判斷這n點能不能組成一個多邊形，並且點都在直角上，如果可以輸出周長，不行輸出-1.

思路：對於橫座標為x的所有點，如果是奇數個點，那麼肯定不行，因為同一橫座標x的點，要能折出去再折回來，必然為偶數個，

代碼：

#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;}