Ultraviolet A 10641-Barisal Stadium (dp + ry)

Link: Ultraviolet A 10641-Barisal Stadium

The surrounding points of the playground are given clockwise, and the positions where the surrounding lights can be built and the cost of building the lights at the place are given. The illumination range is tangent to the boundary of the playground, now a minimum cost is required, and all sides of the playground are exposed.

Solution: dp [I] [j] indicates the minimum cost of edge radiation from vertex I to vertex j. The transfer equation is also well written, as long as there is an equal illumination range that covers I and j, it can be extended.

However, the main problem now is how to find the illumination range of each point. At first, I used the position of the lamp and all the points of the boundary to find the slope. The biggest one was the left boundary, and the smallest one was the right boundary, however, WA later found that there was no scientific basis for this practice, but it was helpless to solve the problem by referring to other people's questions. In this case, first find the center of the playground, and then the edge of each playground. If the center is located on the same side of the edge, it cannot be exposed, no side can be exposed.

Also, when processing the ring, I directly double the array.

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;const int N = 105;const int M = 1005; const int INF = 0x3f3f3f3f;const double eps = 1e-6;const double pi = atan(1.0)*4;struct state {int l, r, val;}s[M];struct point {double x, y;point (double x = 0, double y = 0) { this->x = x; this->y = y;}point operator - (const point &o) const {return point(x - o.x, y - o.y);}double det(const point &o) const {return x * o.y - y * o.x;}}p[N], o;int n, m, dp[N];inline double dis(double x, double y) {return sqrt(x*x+y*y);}inline int sign(double x) {return x < -eps ? -1 : x > eps;}inline double getP(double y, double x) {if (fabs(x) < eps) {return y > 0 ? -pi : pi;}return atan2(y, x);}bool judge (point l, point a, point b) {return sign((l - a).det(b - a) * (o - a).det(b - a)) < 0;}void cat (state& u, double xi, double yi) {bool flag[N];memset(flag, false, sizeof(flag));for (int i = 0; i < n; i++) {if (judge(point(xi, yi), p[i], p[i+1]))flag[i] = true;}if (flag[0] && flag[n-1]) {int l = n-1, r = n;while (flag[l]) u.l = l, l--;while (flag[r-n]) u.r = r, r++;} else {int l = 0, r = n-1;while (!flag[l]) l++;u.l = l;while (!flag[r]) r--;u.r = r;}u.r++;if (u.r < u.l)u.r += n;}void init () {o.x = o.y = 0;for (int i = 0; i < n; i++) {scanf("%lf%lf", &p[i].x, &p[i].y);o.x += p[i].x;o.y += p[i].y;}o.x /= n;o.y /= n;p[n] = p[0];double x, y;int value;scanf("%d", &m);for (int i = 0; i < m; i++) {scanf("%lf%lf%d", &x, &y, &value);cat(s[i], x, y);s[i].val = value;}}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 t = i + j;for (int x = 0; x < m; x++) {if (s[x].l > t)continue;int ad = min(s[x].r, i+n);dp[ad] = min(dp[ad], dp[t]+s[x].val);}}ans = min(ans, dp[i+n]);}if (ans == INF)return false;printf("%d\n", ans);return true;}int main () {while (scanf("%d", &n) == 1 && n) {init ();if (!solve())printf("Impossible.\n");}return 0;}