uva 10641 dp

来源：互联网发布：安卓游戏编程教程编辑：程序博客网时间：2024/06/06 19:23

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 110;const double eps = 1e-6;const int INF = 0x3f3f3f3f;int n, m, dp[maxn], c;bool flag[maxn];struct Point{double x, y;Point(double x = 0, double y = 0): x(x), y(y) {}void read(){scanf("%lf%lf", &x, &y);}};Point P[maxn], o, tmp;struct Q{int l, r, c;};Q q[maxn * 10];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, 0, 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++)for (int k = 0; k < m; k++)if (q[k].l <= i + j){int now = min(i + n, q[k].r + 1);dp[now] = min(dp[now], dp[i + j] + q[k].c);}ans = min(ans, dp[i + n]);}if (ans == INF) return 0;printf("%d\n", ans);return true;}int main(int argc, char const *argv[]){while (~scanf("%d", &n) && n){for (int i = 0; i < n; i++)P[i].read();P[n] = P[0];scanf("%d", &m);for (int i = 0; i < m; i++){tmp.read();scanf("%d", &c);q[i] = tra(tmp, c);}if (!solve()) printf("Impossible.\n");}return 0;}

判断灯能否照到，可以用Onleft判断，O(n) 就够了，

重点是dp，第一维枚举起点，第二维表示i之前的点都被照到的最小花费。

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