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一道演算法題目的解法
有這樣一道演算法題目:144張牌放到若干的盒子中, 每個盒子中放10到40張, 問有多少种放法?(不考慮順序)
分析
由144/10=14.4,144/40=3.6可得, 盒子的數量最少為4, 最多為14.
假設有n個盒子(4<=n<=14),每個盒子中分別放a1, a2, …, an張牌, 由於不考慮順序, 不妨設a1<=a2<= … <=an.
然後可以根據盒子的數量進行窮舉, 從而得出答案。
考慮到14個盒子的情況, 30^14是個很可觀的數字, 為了提高演算法的效率, 需要計算ai(1<=i<=n)的最大和最小值。
容易得出
ai最小的情況是
a1=…=ai, ai+1=…=an=40 min(ai)=(144 – (n-i)*40)/i |
ai最大的情況
a1=min(a1) a2= min(a2)…ai-1= min(ai-1), ai= ai+1=…=an max(ai)=(144-min(a1)-min(a2)-...-min(ai-1))/i |
使用深度優先搜尋法的程式如下
#include <stdio.h> #include <stdlib.h> #define COUNT 144 #define MAXVAL 40 #define MINVAL 10 int count = 0; int** minvals; int** maxvals; void get_min_max_vals () { int i; int j; int k; int min, max; int temp; minvals = (int**) malloc (11 * sizeof(int*)); maxvals = (int**) malloc (11 * sizeof(int*)); for (i = 4; i < 15; i++) { minvals[i - 4] = (int*) malloc (i * sizeof(int)); maxvals[i - 4] = (int*) malloc (i * sizeof(int)); for (j = 0; j < i; j++) { min = (COUNT - MAXVAL * (i - j - 1))/(j + 1); minvals[i - 4][j] = min > MINVAL? min: MINVAL; temp = 0; for (k = 0; k < j; k++) { temp += minvals[i - 4][k]; } max = (COUNT - temp) / (i - j); maxvals[i - 4][j] = max > MAXVAL ? MAXVAL: max; } } } void calc (int cur, int total, int* vals) { int i; int from; int last = COUNT; if (cur == total){ for (i = 0; i < cur; i++) { last -= vals[i]; } if ((last >= vals[cur - 1]) && (last <= MAXVAL)) { vals[cur] = last; count ++; } return; } if (0 == cur) { from = minvals[total - 3][cur]; } else { from = minvals[total - 3][cur] > vals[cur - 1] ? minvals[total - 3][cur]: vals[cur - 1]; } for (i = from; i <= maxvals[total - 3][cur]; i++) { vals[cur] = i; calc (cur + 1, total, vals); } } int main() { int j; int* vals; int totalcount; get_min_max_vals(); for (j = 4; j < 15; j++) { count = 0; vals = (int*) malloc (j * sizeof(int)); calc (0, j - 1, vals); printf ("%d boxes: count=%d/n",j, count); totalcount += count; free(vals); } printf ("total=%d/n", totalcount); for (j = 0; j < 11; j++) { free(minvals[j]); free(maxvals[j]); } free(minvals); free(maxvals); return 0; } |