zoukankan      html  css  js  c++  java
  • NOJ——1628Alex’s Game(III)(DFS+回溯)

    • [1628] Alex’s Game(III)

    • 时间限制: 1000 ms 内存限制: 65535 K
    • 问题描述
    • Alex likes to play with one and zero as you know .

      But Alex does’t have a girlfriend now because his girlfriend is trapped in Alcatraz Island.So Alex wants to rescue her.Now Alex has some information about this place.there are two coast(海岸线) which parallel to each otherand there some castles(城堡) in each cosat.Each coast have n(n<=15) castles,each castle has a id which begins from 1.Each castle just has a undirected road with it's adjacent castle .What's more,He must build exactly k bridges between two coast,and the bridge can only be built between two castles which have the same id and the cost is zero.Now Alex is in the first castle in the first coast and his girlfriend is in the last castle in the second coast ,please calculate the shortest distance Alex needs to walk.

    • 输入
    • First are two integers n (n<=15) and k (1 <= k<=n);
      Next contains two lines ,every line contains n-1 integers 
      which represent the distance between two adjacent castles and the value is less than 100.
    • 输出
    • For each case output the answer represent shortest distance Alex needs to walk.
    • 样例输入
    • 4 3
      4 2 6
      2 10 2
    • 样例输出
    • 6

    感觉跟CF上某一题比较像。借助于DFS的树形DP?n只有15的话爆搜好了

    #include<iostream>
    #include<algorithm>
    #include<cstdlib>
    #include<sstream>
    #include<cstring>
    #include<cstdio>
    #include<string>
    #include<deque>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<set>
    #include<map>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MM(x,y) memset(x,y,sizeof(x))
    typedef pair<int,int> pii;
    typedef long long LL;
    const double PI=acos(-1.0);
    const int N=110;
    int n,k,r;
    int dx[35][35];
    int vis[N];
    void dfs(int now,int sum,int bri)
    {
    	if(now==2*n)
    	{
    		r=min(r,sum);
    		return ;
    	}
    	if(now<0||now>2*n)
    		return ;
    	if(now<=n)
    	{
    		if(!vis[now+1])
    		{
    			vis[now+1]=1;
    			dfs(now+1,sum+dx[now][now+1],bri);
    			vis[now+1]=0;
    		}
    		if(!vis[now+n]&&bri+1<=k)
    		{
    			vis[now+n]=1;
    			dfs(now+n,sum+dx[now][now+n],bri+1);
    			vis[now+n]=0;
    		}
    		
    	}
    	else if(now<=2*n&&now>n)
    	{
    		if(!vis[now+1])
    		{
    			vis[now+1]=1;
    			dfs(now+1,sum+dx[now][now+1],bri);
    			vis[now+1]=0;
    		}
    		if(!vis[now-n]&&bri+1<=k)
    		{
    			vis[now-n]=1;
    			dfs(now-n,sum+dx[now][now-n],bri+1);
    			vis[now-n]=0;
    		}		
    	}
    }
    int main(void)
    {
    	int i,j,d;
    	while (~scanf("%d%d",&n,&k))
    	{
    		MM(dx,0);
    		r=INF;
    		for (i=1; i<=n-1; i++)
    		{
    			scanf("%d",&d);
    			dx[i][i+1]=d;
    			dx[i+1][i]=d;
    		}
    		for (i=1; i<=n-1; i++)
    		{
    			scanf("%d",&d);
    			dx[i+n][i+n+1]=d;
    			dx[i+n+1][i+n]=d;
    		}
    		dfs(1,0,0);
    		printf("%d
    ",r);
    	}
    	return 0;
    }
  • 相关阅读:
    算法-最大公约数
    算法-最大连续子序列和
    iOS开发-Bug锦囊
    iOS开发-简单抽奖
    iOS开发-UIActionSheet简单介绍
    iOS开发-UIActivityIndicatorView简单使用
    iOS开发-UITextView实现PlaceHolder的方式
    iOS开发-Reachability实时检测Wifi,2G/3G/4G/网络状态
    [转]jsPlumb插件做一个模仿viso的可拖拉流程图
    Python 日期和时间
  • 原文地址:https://www.cnblogs.com/Blackops/p/5766299.html
Copyright © 2011-2022 走看看