POJ 1011: Sticks

# include <stdio.h># include <stdlib.h># include <string.h># include <string>int first (int A[], int B[], int size) {for (int i = 1; i <= size; ++ i) {if (B[i] == 0) {B[i] = 1;return i;}}return 0;}int find (int A[], int B[], int size, int groupnum, int ith, int sum) {if (groupnum == ith) {return 1;}for (int i = 1; i <= size; ++ i) {if (B[i] == 0 && (sum + A[i]) < A[0]) {B[i] = 1;if (find (A, B, size, groupnum, ith, sum + A[i]))return 1;elseB[i] = 0;}if (B[i] == 0 && (sum + A[i]) == A[0]) {B[i] = 1;int t = first(A, B, size);if (find (A, B, size, groupnum, ith + 1, A[t]))return 1;else {B[i] = 0;B[t] = 0;return 0;}}if (sum == A[0]) {int t = first(A, B, size);if (find(A, B, size, groupnum, ith + 1, A[t]))return 1;else {B[t] = 0;return 0;}}}return 0;}int go(int A[], int B[], int size, int groupnum) {return 0;}void swap (int &a, int &b) {if (a == b) return;a ^= b ^= a ^= b;}int partition (int A[], int p, int r) {int q = p;for (int i = p + 1; i <= r; ++ i) {if (A[i] > A[p]) {++ q;swap (A[q], A[i]);}}swap(A[q], A[p]);return q;}void qsort (int A[], int p, int r) {while (p < r) {int q = partition(A, p, r);qsort(A, p, q - 1);p = q + 1;}}int main () {int n = 0;scanf ("%d", &n);while (n) {int A[n + 1];int B[n + 1];memset (B, 0, sizeof (B));int sum = 0;for (int i = 1; i <= n; ++ i) {scanf ("%d", &A[i]);sum += A[i];}qsort (A, 1, n);for (int i = sum / A[1]; i >= 1; -- i) {if (sum % i == 0) {memset(B, 0, sizeof(B));A[0] = sum / i;if ( find(A, B, n, i, 1, A[first(A, B, n)]) ) {printf("%d\n", A[0]);break;}}}scanf ("%d", &n);}return 0;}

经典深搜+剪枝

简单来讲就是穷举，但有很多地方可以优化。

1：只需考虑1到(sum / max)个组（循环是反过来写的，因为要求最短）

2:  进行排序

3： 假如有n个组，搜完n-1个组就可以了

。。。