UVA 10641

题目链接：10641 - Barisal Stadium

题意：逆时针给定n个点，在给m个灯，每一个灯有一个花费，要求最小花费使得全部边能被灯照到

思路：用向量叉积推断向量的顺逆时针关系，从而预处理出每一个灯能照到的边，然后因为n个点是环的，所以能够直接扩大两倍，dp时候去枚举起点就可以

状态为dp[i]表示如今照到i条边之前的边所有照亮须要的最小花费

代码：

#include <stdio.h>
#include <string.h>

const double eps = 1e-6;
const int N = 105;
const int M = 1005;
int n, m, i, j, dp[N];
bool flag[N];
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))

struct Point {
	double x, y;
	Point(double x = 0, double y = 0) {
		this->x = x;
		this->y = y;
	}
	void read() {
		scanf("%lf%lf", &x, &y);
	}
} p[N], o;

struct Q {
	int l, r, c;
} q[M];


bool judge(Point p0, Point p1, Point p2) {
	 return ((p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y)) < -eps; 
}

Q tra(Point t, int c) {
	Q ans;
	ans.c = c;
	memset(flag, false, sizeof(flag));
	for (int i = 0; i < n; i++) {
		if (judge(t, p[i], p[i + 1]))
			flag[i] = true;
	}
	if (flag[0] && flag[n - 1]) {
		int l = n - 1, r = n;
		while (flag[l - 1]) l--;
		while (flag[r - n + 1]) r++;
		ans.l = l; ans.r = r;
		
	} else {
		int l = 0, r = n - 1;
		while (!flag[l]) l++;
		while (!flag[r]) r--;
		ans.l = l; ans.r = r;
	}
	if (ans.r < ans.l) ans.r += n;
	return ans;
}

bool solve() {
	int ans = INF;
	for (int i = 0; i < n; i++) {
		memset(dp, INF, sizeof(dp));
		dp[i] = 0;
		for (int j = 0; j < n; j++) {
			int r = i + j;
			for (int k = 0; k < m; k++) {
				if (q[k].l > r) continue;
				int now = min(i + n, q[k].r + 1);
				dp[now] = min(dp[now], dp[r] + q[k].c);
			}
		}
		ans = min(ans, dp[i + n]);
	}
	if (ans == INF) return false;
	printf("%d
", ans);
	return true;
}

int main() {
	while (~scanf("%d", &n) && n) {
		for (i = 0; i < n; i++) p[i].read();
		p[n] = p[0];
		scanf("%d", &m);
		Point tmp;
		int c;
		for (i = 0; i < m; i++) {
			tmp.read();
			scanf("%d", &c);
			q[i] = tra(tmp, c);
		}
		if (!solve()) printf("Impossible.
");
	}
	return 0;
}