zoukankan      html  css  js  c++  java
  • HDU 4277 USACO ORZ 第37届ACM/ICPC长春赛区网络赛1011题(搜索)

    USACO ORZ

    Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1099    Accepted Submission(s): 398


    Problem Description
    Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.
    I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N fence segments and must arrange them into a triangular pasture. Ms. Hei must use all the rails to create three sides of non-zero length. Calculating the number of different kinds of pastures, she can build that enclosed with all fence segments.
    Two pastures look different if at least one side of both pastures has different lengths, and each pasture should not be degeneration.
     
    Input
    The first line is an integer T(T<=15) indicating the number of test cases.
    The first line of each test case contains an integer N. (1 <= N <= 15)
    The next line contains N integers li indicating the length of each fence segment. (1 <= li <= 10000)
     
    Output
    For each test case, output one integer indicating the number of different pastures.
     
    Sample Input
    1 3 2 3 4
     
    Sample Output
    1
     
    Source
     
    Recommend
    liuyiding
     
     
     
    就是搜索,在长春网络赛上做得还是比较顺畅的题目了。
     
    做搜索题目。。。还是稍微好点了。
     
    我用map判重的,而且要注意几个地方要剪枝,否则会TLE的。
     
    #include<stdio.h>
    #include<math.h>
    #include<iostream>
    #include<algorithm>
    #include<map>
    using namespace std;
    const int MAXN=20;
    int a[MAXN];
    map <long long,int> mp;
    int tol;
    int n;
    int sum;
    void dfs(int l1,int l2,int l3,int i)
    {
        if(l1>l2)swap(l1,l2);
        if(l1>l3)swap(l1,l3);
        if(l2>l3)swap(l2,l3);
        if(l3*2>sum)return;
        if(i==n)
        {
            if(l1+l2>l3)
            {
                long long t=l1*150000*150000+l2*150000+l3;
                if(mp[t]==0)
                {
                    mp[t]=1;
                    tol++;
                }
            }
            return;
        }
        int t11=l1+a[i+1];
        int t12=l2;
        int t13=l3;
        if(t11>t12)swap(t11,t12);
        if(t11>t13)swap(t11,t13);
        if(t12>t13)swap(t12,t13);
        dfs(t11,t12,t13,i+1);
        int t21=l1;
        int t22=l2+a[i+1];
        int t23=l3;
        if(t21>t22)swap(t21,t22);
        if(t21>t23)swap(t21,t23);
        if(t22>t23)swap(t22,t23);
        if(t11!=t22 || t12!=t22 ||t13!=t23)dfs(t21,t22,t23,i+1);
        int t31=l1;
        int t32=l2;
        int t33=l3+a[i+1];
        if(t31>t32)swap(t31,t32);
        if(t31>t33)swap(t31,t33);
        if(t32>t33)swap(t32,t33);
        if(t31!=t11||t32!=t12||t33!=t13)
          if(t31!=t21||t32!=t22||t33!=t23)
             dfs(t31,t32,t33,i+1);
    }
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&n);
            sum=0;
            for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];}
            mp.clear();
            tol=0;
            dfs(0,0,0,0);
            printf("%d\n",tol);
        }
        return 0;
    }
  • 相关阅读:
    android技能树
    BitmapFactory 读取图片方法总结
    如何利用SecureCRT连接Ubuntu12.0.4
    Ubuntu 12.04如何从登录界面登录root
    ubuntu中vi在编辑状态下方向键不能用的解决
    在ubuntu12.0.4上搭建samba服务器以实现文件共享
    怎么利用ultraISO对一个文件夹制作ISO镜像
    VMware网络选项分析
    在vmware 6.5+ubuntu12.04上安装VMware tools出现问题的分析
    在ubuntu中获得root权限
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2679034.html
Copyright © 2011-2022 走看看