ACM 算法艺术与信息学竞赛 1.2.1盒子里面的汽球

盒子里面的汽球

题目大意我就不多说了，直看一下刘汝佳的书。这里给出链接，与我出错的总结

盒子里的汽球

Problem 1515 Balloons in a Box

Accept: 171    Submit: 919
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

You must write a program that simulates placing spherical balloons into a rectangular box.

The simulation scenario is as follows. Imagine that you are given a rectangular box and a set of points inside the box. Each point represents a position where you might place a balloon. To place a balloon at a point, center it at the point and inflate the balloon until it touches a side of the box or a previously placed balloon. You may use the points in any order you like, and need not use every point. Your objective is to place balloons in the box in an order that maximizes the total volume occupied by the balloons.

You are required to calculate the volume within the box that is not enclosed by the balloons.

All integer will be in [-1000, 1000].

Input

The input consists of several test cases. The first line of each test case contains a single integer n that indicates the number of points in the set (n ≤ 6). The second line contains three integers that represent the (x, y, z) integer coordinates of a corner of the box, and the third line contains the (x, y, z) integer coordinates of the opposite corner of the box. The next n lines of the test case contain three integers each, representing the (x, y, z) coordinates of the points in the set. The box has non-zero length in each dimension and its sides are parallel to the coordinate axes.

Output

For each test case print one line which indicates the volume of the box not occupied by balloons. Round the volume to the nearest integer.

Sample Input

2 0 0 0 10 10 10 3 3 3 7 7 7

Sample Output

774

Source

FZU 2007 ICPC Qualification Round I

出错总结

1、用GNU C 提交，但是加了//为注释

2、在查看别的点的时候，如果别的点的半径为0，可需要直接跳过

for (j = 0; j < n; ++j){iter = ret[j]; mind = INF;for (k = 0; k < n; ++k){if (k == iter) continue;  ->改为 if (k == iter || radius[k] == 0) continue;d = D[iter][k] - radius[k];mind = min(d, mind);}radius[iter] = min(mind, defined[iter]);}

3、可能福州里面使用的编译器版本比较老，我用while语句会出错，但是改成for则没有问题。

void _get_number(int v, const int n){int idx = 1, now_value = n, iter = n, pos, cnt = -1, j;memset(rec, 0, sizeof(rec));memset(ret, 0, sizeof(ret));do {rec[idx] = v % (idx + 1);v /= ++idx;} while (v);while (--iter && --now_value)  ->改为 for (iter = now_value = MAXN -1; iter > 0; --iter, --now_value){pos = rec[iter];for (cnt = -1, j = n - 1; j >= 0 && cnt != pos; --j) cnt += 0 == ret[j];ret[j + 1] = now_value;}}

4、此外，坐标值最好用double来表示。（让我错了无数次，郁闷）

5、四舍五入的处理用printf("%.0lf\n", V);可以实现四舍五入的效果。

附上代码吧，这里用的是枚举的思路来处理的。不知道有没有更好的思路

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#define  MM 100const double PI = 3.1415926535;const double INF = 2147364735.0;#define  min(a, b)  ((a) > (b) ? (b) : (a))#define  max(a, b)  ((a) > (b) ? (a) : (b))#define  abso(a) ((a) > 0 ? (a) : 0 - (a))const int frac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};typedef struct _node{double x, y, z;}node;node point[MM], from, to;double D[MM][MM];double R[MM];int rec[MM];int ret[MM];void _get_number(const int v, const int MAXN){int idx = 1, iter = MAXN, pos, cont, j;int value = v;memset(rec, 0, sizeof(rec));memset(ret, 0, sizeof(ret));do{rec[idx] = value % (idx + 1);value /= ++idx;} while (value);for (iter = MAXN - 1; iter > 0; --iter){pos = rec[iter];for (cont = -1, j = MAXN - 1; j >= 0; --j){if (0 == ret[j]){++cont;if (pos == cont) { ret[j] = iter; break;}}}}}double _dist(int i, int j){double a = point[i].x - point[j].x;double b = point[i].y - point[j].y;double c = point[i].z - point[j].z;return sqrt(a * a + b * b + c * c);}double _min_dist(int i){double t;double f = point[i].x - from.x, s = to.x - point[i].x; f = abso(f); s = abso(s); t = min(f, s);f = point[i].y - from.y; s = to.y - point[i].y; f = abso(f); s = abso(s); t = min(f, t); t = min(s, t);f = point[i].z - from.z; s = to.z - point[i].z; f = abso(f); s = abso(s); t = min(f, t); t = min(s, t);return t;}int main(void){int n, i, j, k, next, iter;double V, minR, sumV, maxV, d;while (scanf("%d", &n) != EOF){scanf("%lf%lf%lf%lf%lf%lf", &(from.x), &(from.y), &(from.z), &(to.x), &(to.y), &(to.z));V = (to.x - from.x) * (to.y - from.y) * (to.z - from.z); V = abso(V);for (i = 0; i < n; ++i) scanf("%lf%lf%lf", &(point[i].x), &(point[i].y), &(point[i].z));for (i = 0; i < n; ++i) for (D[i][i] = _min_dist(i), j = i + 1; j < n; ++j) D[i][j] = D[j][i] = _dist(i, j);for (maxV = next = 0; next < frac[n]; ++next){memset(R, 0, sizeof(R));_get_number(next, n);for (sumV = i = 0; i < n; ++i){iter = ret[i]; minR = INF;for (k = 0; k < n; ++k){if (k != iter && R[k] > 0){d = D[iter][k] - R[k];if (d <= 0) { minR = 0; break;}else minR = min(minR, d);}}minR = min(minR, D[iter][iter]); R[iter] = minR;sumV += 4.0 * PI * minR * minR * minR / 3.0;}maxV = max(maxV, sumV);}printf("%.0lf\n", V - maxV);}return 0;}