ZOJ 1204:(DFS)

输出一串数所能构成的所有等式,按照加数从大到小,算式从长到短排列

一开始因为无法按照要求输出纠结了好一阵~~最后看了大神的代码才发现只需要每次dfs规定长度就好了~

#include"cstdio"
#include"cstring"
#include"algorithm"
#define MAXN 40
using namespace std;
int num[MAXN],n,front,rear,ok;
int stack[MAXN];
int cal()
{   int sum=0;
    for(int i=front;i<rear;i++)
        sum+=stack[i];
    return sum;
}
void display()
{   for(int i=front;i<rear-1;i++)
        printf("%d+",stack[i]);
    printf("%d=%d
",stack[rear-1],cal());
}
bool find(int x)
{   int low=0,high=n-1,mid;
    while(high>=low)
    {   mid=low+(high-low)/2;
        if(num[mid]==x) return true;
        else if(num[mid]>x) high=mid-1;
        else low=mid+1;
    }
    return false;
}
void dfs(int mark,int len)
{   for(int i=mark+1;i<n;i++)
    {   if(i<n&&cal()+num[i]<=num[n-1])
        {   stack[rear++]=num[i];
            if(rear-front==len&&find(cal())) {display();ok=0;}
            else dfs(i,len);
            rear--;
        }
    }
}
int main()
{   int t;
    scanf("%d",&t);
    while(t--)
    {   ok=1;
        memset(stack,0,sizeof(stack));
        memset(num,0,sizeof(num));
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",&num[i]);
        sort(num,num+n);
        for(int len=2;len<=n;len++)
        for(int i=0;i<n;i++) {front=rear=0;stack[rear++]=num[i];dfs(i,len);}
        if(ok) printf("Can't find any equations.

");
        else printf("
");
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/luxiaoming/p/4671058.html