poj1011 搜索+剪枝

DFS+剪枝 POJ2362的强化版,重点在于剪枝 令InitLen为所求的最短原始棒长,maxlen为给定的棒子堆中最长的棒子,sumlen为这堆棒子的长度之和,那么InitLen必定在范围[maxlen,sumlen]中 根据棒子的灵活度(棒子越长,灵活度越低) DFS前先对所有棒子降序排序 剪枝: 1、 由于所有原始棒子等长,那么必有sumlen%Initlen==0; 2、 若能在[maxlen,sumlen-InitLen]找到最短的InitLen,该InitLen必也是[maxlen,sumlen]的最短;若不能在[maxlen,sumlen-InitLen]找到最短的InitLen,则必有InitLen=sumlen; 3、 由于所有棒子已降序排序,在DFS时,若某根棒子不合适,则跳过其后面所有与它等长的棒子; 4、 最重要的剪枝:对于某个目标InitLen,在每次构建新的长度为InitLen的原始棒时,检查新棒的第一根stick[i],若在搜索完所有stick[]后都无法组合,则说明stick[i]无法在当前组合方式下组合,不用往下搜索(往下搜索会令stick[i]被舍弃),直接返回上一层 

#include<iostream>
#include<algorithm>
using namespace std;

int cmp( const void* a, const void* b )
{
    return *( int* )b - *( int* )a;
}

int n;
bool dfs( int* stick, bool* visit, int len, int InitLen, int s, int num )
{
    if ( num == n )
        return true;

    int sample = -1;
    for( int i = s; i<n; i++ )
    {
        if( visit [i] || stick[i] == sample )
            continue;
        visit[i] = true;
        if( len + stick[i] < InitLen )
        {
            if( dfs( stick, visit, len+stick[i], InitLen, i, num+1))
                return true;
            else
                sample = stick[i];
        }
        else if( len + stick[i] == InitLen )
        {
            if( dfs( stick, visit, 0, InitLen, 0, num+1 ))
                return true;
            else sample = stick[i];
        }
        visit[i] = false;

        if( len == 0 )
            break;

    }
    return false;
}

int main()
{
    while( cin >> n && n )
    {
        int* stick = new int[n];
        bool* visit = new bool[n];
        int sumlen = 0;
        int i;
        for( i = 0; i<n; i++ )
        {
            cin >> stick[i];
            sumlen+=stick[i];
            visit[i] = false;
        }

        qsort( stick, n, sizeof(stick), cmp );
        int  maxlen = stick[0];

        bool flag = false;

        for ( int InitLen = maxlen;InitLen <= sumlen - InitLen; InitLen++)
        {
            if( !( sumlen%InitLen) && dfs( stick, visit, 0, InitLen, 0, 0 ))
            {

                cout<< InitLen <<endl;
                flag = true;
                break;
            }
        }
        if( !flag )
            cout<< sumlen << endl;

        delete stick;
        delete visit;
    }
    return 0;
}

 

posted @ 2015-03-29 14:33  peerslee2  阅读(121)  评论(0编辑  收藏  举报