2013多校九1003-HDU4688-DP+计算几何

不讲题意了。

思路：

状态就是DP(i, j)表示i前面的切好了，j后面的切好了，求i到j这一节的答案。

DP思想就是这样。

计算几何真心不会写。计算几何是抄师傅的。

const int N = 222;const double eps = 1e-10;int vis[N][N];//这个状态求过double res[N][N];double to[N][2];//向左向右距离边缘的距离double dd[N][N];//两线段相交，后一个线段距离交点的距离bool lx[N][N];int n;double a, b;double MAX;struct Point {    double x, y;    Point() {}    Point(double _x, double _y) : x(_x), y(_y) {}    Point operator -(const Point &p) const {        return Point(x - p.x, y - p.y);    }    double xMul(const Point &p) const {        return x * p.y - y * p.x;    }} p[N];inline double dis(const Point &a, const Point &b) {    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}struct Line {    Point u, v;    double w;    Line() {}    Line(const Point &_u, const Point &_v) : u(_u), v(_v) {        w = dis(_u, _v);    }} line[N];inline bool parallel(const Line &a, const Line &b) {    return ZERO((a.v - a.u).xMul(b.v - b.u));}inline Point inter(const Line &a, const Line &b) {    double s1 = (b.v - a.u).xMul(b.u - a.u), s2 = (b.u - a.v).xMul(b.v - a.v);    return Point((a.u.x * s2 + a.v.x * s1) / (s2 + s1), (a.u.y * s2 + a.v.y * s1) / (s2 + s1));}void init(){    int i, j;    Point o, q[] = {Point(0, 0), Point(a, 0), Point(a, b), Point(0, b), Point(0, 0)};    double l, r;    for (i = 0; i < 2*n ; ++i) {        for (j = 0; j < 2*n ; ++j) {            if (j == i) continue;            if (parallel(line[i], line[j])) {                lx[i][j] = true; dd[i][j] = MAX;            } else {                o = inter(line[i], line[j]);                l = dis(o, p[i]); r = dis(o, p[i + 1]);                if (l < r) {                    lx[i][j] = true; dd[i][j] = l;                } else {                    lx[i][j] = false; dd[i][j] = r;                }            }        }    }    for (i = 0; i < 2*n; ++i) {        to[i][0] = to[i][1] = MAX;        for (j = 0; j < 4; ++j) {            if (parallel(line[i], Line(q[j], q[j + 1]))) continue;            o = inter(line[i], Line(q[j], q[j + 1]));            l = dis(o, p[i]); r = dis(o, p[i + 1]);            if (l < r) to[i][0] = min(to[i][0], l);            else to[i][1] = min(to[i][1], r);        }            //cout<<i<<' '<<p[i].x<<' '<<p[i].y<<' '<<to[i][0]<<' '<<to[i][1]<<endl;    }}double DP(int i, int j){    if(i==j) return 0;    if(vis[i][j]) return res[i][j];    vis[i][j] = 1;    double ans = MAX;    for(int k=i; k<j; ++k){        double ll = to[k][0];//点k到边界的距离        double rr = to[k][1];//点k+1到边界的距离        if(lx[k][i-1]) ll = min(ll, dd[k][i-1]);//如果k这头与i-1这条线段有交点        else rr = min(rr, dd[k][i-1]);        if (lx[k][j]) ll = min(ll, dd[k][j]);        else rr = min(rr, dd[k][j]);        ans = min(ans, DP(i, k) + DP(k+1, j) + ll + rr + line[k].w );    }    return res[i][j] = ans;}int main(){    while(scanf("%d%lf%lf", &n, &a, &b)!=EOF){        for (int i = 0; i < n; ++i) scanf("%lf%lf", &p[i].x, &p[i].y);        for (int i = n; i <= 2 * n; ++i) p[i] = p[i - n];        for (int i = 0; i < 2 * n; ++i) line[i] = Line(p[i], p[i + 1]);        MAX = (a+b)*1000.0;        init();        memset(vis, 0, sizeof(vis));        double ans = MAX;        for (int i = 1; i <= n; ++i) {//枚举第一刀            ans = min(ans, DP(i, i + n - 1) + to[i - 1][0] + to[i - 1][1] + line[i - 1].w);        }        printf("%lf\n", ans);    }    return 0;}