UVA 10641 - Barisal Stadium（DP + 几何）

题目链接：10641 - Barisal Stadium

题意：逆时针给定n个点，在给m个灯，每个灯有一个花费，要求最小花费使得所有边能被灯照到
思路：用向量叉积判断向量的顺逆时针关系，从而预处理出每个灯能照到的边，然后由于n个点是环的，所以可以直接扩大两倍，dp时候去枚举起点即可
状态为dp[i]表示现在照到i条边之前的边全部照亮需要的最小花费
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
;
bool flag
;
#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
, 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\n", ans);
return true;
}

int main() {
while (~scanf("%d", &n) && n) {
for (i = 0; i < n; i++) p[i].read();
p
= 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.\n");
}
return 0;
}