动态规划 ：POJ 1014 Dividing

#include <stdio.h>#include <string.h>int array[7] = {0};char flag[60001] = {0};int main(){int sum = 0, n = 0;while (1){n++;scanf("%d %d %d %d %d %d", &array[1], &array[2], &array[3], &array[4], &array[5], &array[6]);sum = 1*array[1] + 2*array[2] + 3*array[3] + 4*array[4] + 5*array[5] + 6*array[6];if (sum == 0){break;}if (sum % 2 == 1){printf("Collection #%d:/nCan't be divided./n/n", n);continue;}flag[0] = 1;int i, j, k;for (i=1; i<=6; ++i){if (array[i] == 0){continue;}for (j=sum/2; j>=0; --j){if (flag[j] == 1){for (k=1; k<=array[i]; ++k){if (j+k*i > sum/2 || flag[j+k*i] == 1){break;}flag[j+k*i] = 1;if (flag[sum/2] == 1){break;}}}}}if (flag[sum/2] == 1){printf("Collection #%d:/nCan be divided./n/n", n);}else{printf("Collection #%d:/nCan't be divided./n/n", n);}memset(flag, 0, 60001);}return 0;}

思路：是否能均分，可以转化为是否能够拿出value为sum/2的石子。flag[i]如果为1，则表示能够拿出value为i的石子。所以问题的最终解即为：flag[sum/2]是否为1。
如果存在value为K的石子，且能够拿出value为i的石子，即flag[i] = 1，则肯定也能拿出value为i+K的石子，即flag[i+K]肯定也为1。所以，我们的初始状态是flag[0] = 1，要从它推导出flag[sum/2]是否为1。