UVALive 4651(DP)

来源：互联网发布：怪医黑杰克知乎编辑：程序博客网时间：2024/06/05 20:43

题意是给你ｎ个点，你需要从第一个到第ｎ个点连一条折线，折线中间按顺序经过若干个点，并且折线外的点离这条折线的最近距离需要小于Ｄ。需要求这条折线的最小长度。

DP[i][j]表示到第ｉ个点为止折线上有ｊ条线段的最短长度，那么DP[k][j+1] = min (dp[k][j+1], dp[i][j]+distance (i, j)), 其中ｋ>ｉ，并且在线段ｉｋ之间的每个点都满足点到线段的距离小于等于Ｄ。

#include <bits/stdc++.h>using namespace std;#define maxn 111#define INF 1e20double dp[maxn][maxn];int n;double x, y, d;double dis[maxn][maxn];bool ok[maxn][maxn];struct point {    double x, y;    point (double xx = 0, double yy = 0) : x(xx), y(yy) {}}p[maxn];point operator + (point a, point b) {        return point (a.x+b.x, a.y+b.y);    }point operator - (point a, point b){    return point (a.x-b.x, a.y-b.y);}bool operator < (point a, point b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}const double eps = 1e-10;int dcmp (double x) {    if (fabs (x) < eps)        return 0;    else        return x < 0 ? -1 : 1;}bool operator == (const point &a, const point &b) {    return dcmp (a.x-b.x) == 0 && dcmp (a.y-b.y) == 0;}double Dot (point a, point b) {    return a.x*b.x + a.y*b.y;}double Length (point a) {    return sqrt (Dot (a, a));}double Cross (point a, point b) {    return a.x*b.y - a.y*b.x;}double DistanceToSegment (point p, point a, point b) {    if (a == b)        return Length (p-a);    point v1 = b-a, v2 = p-a, v3 = p-b;    if (dcmp (Dot (v1, v2)) < 0)        return Length (v2);    else if (dcmp (Dot (v1, v3)) > 0)        return Length (v3);    else return fabs (Cross (v1, v2)) / Length (v1);}double get_dis (int i, int j) {    double x1 = p[i].x, y1 = p[i].y, x2 = p[j].x, y2 = p[j].y;    return sqrt ((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}void legal () {    for (int i = 1; i <= n; i++) {        for (int j = i+1; j <= n; j++) {            ok[i][j] = 1;            for (int k = i+1; k < j; k++) {                double dd = DistanceToSegment (p[k], p[i], p[j]);                if (dd > d) {                    ok[i][j] = 0;                    break;                }            }        }    }}int main () {    //freopen ("in", "r", stdin);    while (cin >> n >> d && n+d) {        for (int i = 1; i <= n; i++) {            cin >> p[i].x >> p[i].y;        }        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= n; j++) {                dp[i][j] = INF;            }        }        for (int i = 1; i <= n; i++) {            for (int j = i+1; j <= n; j++)                dis[i][j] = dis[j][i] = get_dis (i, j);        }        legal ();        for (int i = 2; i <= n; i++) {            if (ok[1][i]) {                dp[i][1] = dis[1][i];            }        }        for (int i = 1; i <= n; i++) {            for (int j = 1; j < i; j++) {                for (int k = i+1; k <= n; k++) {                    if (ok[i][k] && dp[i][j] != INF) {                        dp[k][j+1] = min (dp[k][j+1], dp[i][j]+dis[i][k]);                    }                }            }        }        double ans = INF;        for (int i = 1; i < n; i++) {            ans = min (ans, dp[n][i]);        }        printf ("%.2lf\n", ans);    }    return 0;}

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