DFS + 剪枝 —— POJ 1011

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Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 125013 Accepted: 29107

Description

George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.

Input

The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

Output

The output should contains the smallest possible length of original sticks, one per line.

Sample Input

95 2 1 5 2 1 5 2 141 2 3 40

Sample Output

65

POJ 2362 的加强版

题意：给一个数n， 然后给出n个数，每个数代表一根木棍长度，求这n根木棍恰好可以组成k根长度都为len的木棍。求len的最小值。

思路：搜索枚举 + 剪枝。

1、降序排序。

2、需要组成的长度len必须为总长sum的公约数。

3、枚举区间：最大木棍长度 ～ 总木棍长度。

4、在尝试组成一根长为len的木棍时，从第一根子木棍开始，如果失败了，那这根子木棍在后面肯定也用不上，所以直接返回失败。

5、在尝试组成一根长为len的木棍时，如进行到 11     8     ...   若 8  失败了，如果后面的还是8，肯定也不会成功，那直接跳过。

6、如果需要组k根长为len的木棍，那完成k-1次匹配后应该就可以了的，最后一次应该也可以匹配。不过提交却是TLE，可能是我考虑欠妥。

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#define M 70using namespace std;int a[M];int vis[M];int n;int Dfs(int cnt, int beg, int t, int key_t, int sum, int key_len){if(sum == key_len){ //组成一根长为key_len的棍子t++;if(t == key_t) return 1; //棍子数为sum / key_len，成功。按道理这里改为 t == key_t - 1 似乎也是正确的，提交却是TLEsum = 0;for(beg=0; beg<n; beg++) //beg为下一根棍子的起始位置下标if(!vis[beg]) break;cnt = beg;}int i, flag = 0;for(i=cnt; i<n; i++){if(sum + a[i] <= key_len && !vis[i]){vis[i] = 1;sum += a[i];flag = Dfs(i+1, beg, t, key_t, sum, key_len);if(1 == flag) return 1;sum -= a[i];vis[i] = 0;while(a[i] == a[i+1]) i++;}if(!vis[beg]) return 0;}return 0;}bool cmp(int a, int b){return a > b;}int main(){//freopen("in.txt", "r", stdin);while(~scanf("%d", &n), n){memset(vis, 0, sizeof(vis));int i;int sum = 0;for(i=0; i<n; i++){scanf("%d", &a[i]);sum += a[i];}sort(a, a+n, cmp);int ans, key_len;int ok = 0;for(ans=a[0]; ans<= sum; ans++){if(sum % ans) continue;key_len = sum / ans;ok = Dfs(0, 0, 0, key_len, 0, ans);if(ok) break;}printf("%d\n", ans);}return 0;}