hdu1455 && poj1011 Sticks（深度优先搜索 DFS 经典剪枝 详解）

这道题在两个OJ上都有

POJ地址

http://poj.org/problem?id=1011

HDU地址

http://acm.hdu.edu.cn/showproblem.php?pid=1455

Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 129296		Accepted: 30304

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

9
5 2 1 5 2 1 5 2 1
4
1 2 3 4
0

Sample Output

6
5

题目大意

给你一些短的木棒拼成相同长度的长的木棒，问你所能拼成的木棒最短的长度

思路

用到多次的剪枝；

剪枝一：棍子L的长度一定会被所有棍子总和ntot整除，不能整除的减掉；

剪枝二：用lasttickno记录上次的最后一根棍子的序号，下次从lasttickno+1开始；

剪枝三：如果某次拼接选择长度为S 的木棒，导致最终 失败，则在同一位置尝试下一根木棒时，要跳 过所有长度为S 的木棒 ；

剪枝四： 如果在不替换第一根木棒的情况下怎么都无法成 功，那么就要推翻第i-1根棍子的拼法 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int N,l,s;
int vis[65],len[65];
int sum;
int laststickno;
int dfs(int r,int m)//r为所剩的棍子的个数， m表示当前正在拼的棍子和 l 比还缺的长度 
{
    if(r==0&&m==0)
        return true;
    if(m==0)
            m=l;
    int startno=0;
    if(m!=l)
        startno=laststickno+1;//剪枝2
    for(int i=startno;i<s;i++){
        if(!vis[i]&&len[i]<=m){
            if(i>0){
                if(vis[i-1]==false&&len[i]==len[i-1])//剪枝3
                        continue;

            }
            vis[i]=1;
            laststickno=i;
            if(dfs(r-1,m-len[i]))
                return true;
            else{
                vis[i]=0;
                if(len[i]==l||m==l)
                    return false;
            }
        }
    }
    return false;
}
int main()
{
    while(cin>>s&&s)
    {
        laststickno=s-1;
        int ntot=0;
        memset(vis,0,sizeof(vis));
        memset(len,0,sizeof(len));
        for(int i=0;i<s;i++){

            scanf("%d",&len[i]);
            ntot+=len[i];
        }
        sort(len,len+s);
        for(l=len[0];l<=ntot/2;l++){
            if(ntot%l)
                continue;
            memset(vis,0,sizeof(vis));
            if(dfs(s,l)){
                cout<<l<<endl;
                break;
            }
        }
        if(l>ntot/2)
            cout<<ntot<<endl;//剪枝4
    }
    return 0;

}