NYOJ-325-zb的生日(典型搜索)

题目链接：
http://acm.nyist.net/JudgeOnline/problem.php?pid=325

这个题用搜索和01背包都可以做，但是01背包超时了( ▼-▼ )

1.
常规搜索（时间有点长）：

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
int minn,sum;
int a[100];
void dfs(int m,int i,int cur)
{
    if(i<m)
    {
        int tmp=sum-cur*2;  //tmp的绝对值表示差值；
        if(tmp<0)tmp=-tmp;
        if(tmp<minn)minn=tmp;
        dfs(m,i+1,cur+a[i+1]);
        dfs(m,i+1,cur);
    }
    return ;
}
int main()
{
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        sum=0;
        memset(a,0,sizeof(a));
        for(int i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        minn=sum;
        dfs(m,0,a[0]);
        printf("%d\n",minn);
    }
    return 0;
}        

2.
高效搜索：

#include<iostream>
#include<cstring>
#include<cstdio> 
using namespace std;
#define max(a,b) a>b?a:b

int V, ans, n, w[21], sum[21];

void dfs(int i,int cnt)  
{  
    if(i == 0)  
    {  
        ans = max(ans,cnt);
        return ;  
    } 

    if(ans == V || cnt+sum[i] <= ans)       
        return ;

    if(cnt+w[i] <= V)  
        dfs(i-1,cnt+w[i]);//前一个放
    dfs(i-1,cnt);  //前一个不放
}

int main()  
{  
    while(scanf("%d",&n) != EOF)  
    {  
        ans = 0;
        memset(w, sizeof(w), 0);
        memset(sum, sizeof(sum), 0);
        for(int i=1;i<=n;i++)  
        {  
            scanf("%d",&w[i]);  
            sum[i] = sum[i-1] + w[i];
        }  
        V = sum[n]/2; //只搜一半
        dfs(n,0);  
        printf("%d\n",sum[n]-2*ans);  
    }  
    return 0;  
}

3.
01背包做法（超时）：

#include<stdio.h> 
#include<string.h> 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int sum;
int b[20005];
int a[100];
int main()
{
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        sum=0;
        memset(b,0,sizeof(b));
        for(int i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        for(int i=0;i<m;i++)
        {
            for(int j=sum/2;j>=a[i];j--)//用一半容积
            {
                b[j]=max(b[j],b[j-a[i]]+a[i]);
            }
        }
        printf("%d\n",sum-2*b[sum/2]);
    }
    return 0;
}        

仅代表个人观点，欢迎交流探讨，勿喷~~

PhotoBy:WLOP

http://weibo.com/wlop