• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            posts - 195,  comments - 30,  trackbacks - 0
            I have my friends to visit. For some reason, I can only visit some of my friends. So I want see my friends as many as possible. Thus I must choose the longest way. Your goal is to help me developing a program that computes the length of the longest path that can be constructed in a given graph from a given starting point (My residence). You can assume that the graph has no cycles (there is no path from any node to itself), so I will reach my destination in a finite time. In the same line of reasoning, nodes are not considered directly connected to themselves.

            Input

            The input consists of a number of cases. The first line on each case contains a positive number n (1 < n <= 100) that specifies the number of points in the graph. A value of n = 0 indicates the end of the input. After this, a second number s is provided, indicating the starting point in my journey (1 <= s <= n). Then, you are given a list of pairs of places p and q, one pair per line. The pair "p q" indicates that I can visit q after p. A pair of zeros ("0 0") indicates the end of the case. As mentioned before, you can assume that the graphs provided will not be cyclic.

            Output

            For each test case you have to find the length of the longest path that begins at the starting place. You also have to print the number of the final place of such longest path. If there are several paths of maximum length, print the final place with smallest number.

            Print a new line after each test case.

            Sample Input

            2
            1
            1 2
            0 0
            5
            3
            1 2
            3 5
            3 1
            2 4
            4 5
            0 0
            5
            5
            5 1
            5 2
            5 3
            5 4
            4 1
            4 2
            0 0
            0
            

            Sample Output

            Case 1: The longest path from 1 has length 1, finishing at 2.
            Case 2: The longest path from 3 has length 4, finishing at 5.
            Case 3: The longest path from 5 has length 2, finishing at 1.
            提交了10次,AC 3次,超時4次,wa 3次。
            很無語。
            應(yīng)該是動態(tài)規(guī)劃最好,但是我不是很熟,用了搜索。以下是超時的代碼
            #include<iostream>
            #include<cstdlib>
            using namespace std;
            int path[101][101];
            int mark[101];
            int len[101];//
            int end[101];
            int dfs(int start,int num)//返回從當(dāng)前點出發(fā)的最大長度
            {
            if(mark[start]==1)return len[start];
            mark[start]=1;
            end[start]=start;
            for(int i=1;i<=num;i++)
            {
            if(path[start][i])
            {
            if(len[start]<dfs(i,num)+1)
            {
            len[start]=len[i]+1;
            end[start]=end[i];
            }
            else
            if(len[start]==len[i]+1&&end[start]>end[i])
            end[start]=end[i];
            }
            }
            mark[start]=0;//這句堅決不需要 
            return len[start];
            }
            int main()
            {
            // freopen("s.txt","r",stdin);
            //freopen("key.txt","w",stdout);
            int j,k,turn=0;
            int start,num;
            while(cin>>num)
            {
            turn++;
            if(num==0)break;
            memset(path,0,sizeof(path));
            memset(mark,0,sizeof(mark));
            memset(len,0,sizeof(len));
            memset(end,0,sizeof(end));
            cin>>start;
            while(cin>>j>>k)
            {
            if(j==0)break;
            path[j][k]=1;
            }
            cout<<"Case "<<turn<<": The longest path from "<<start<<" has length "<<dfs(start,num)<<", finishing at "<<end[start]<<"."<<endl<<endl;
            }
            //system("PAUSE");
            return   0;
            }
            以下是ac的代碼
            #include<iostream>
            #include<cstdlib>
            using namespace std;
            int path[102][102];
            int mark[102], len[102],end[102];
            int dfs(int start,int num)//返回從當(dāng)前點出發(fā)的最大長度
            {
            if(mark[start]==1)return len[start];
            mark[start]=1;
            end[start]=start;
            len[start]=0;
            int i,t;
            for( i=1;i<=num;i++)
            {
            if(path[start][i])
            {
            t=dfs(i,num)+1;
            if(t>len[start])
            {
            len[start]=t;
            end[start]=end[i];
            }
            else
            if(len[start]==t)
            {
            if(end[start]>end[i])
            end[start]=end[i];
            }
            }
            }
            return len[start];
            }
            int main()
            {
            //freopen("s.txt","r",stdin);
            //freopen("key.txt","w",stdout);
            int j,k,turn=0;
            int start,num;
            while(cin>>num,num)
            {
            turn++;
            memset(path,0,sizeof(path));
            memset(mark,0,sizeof(mark));
            memset(len,0,sizeof(len));
            memset(end,0,sizeof(end));
            cin>>start;
            while(cin>>j>>k,j||k)
            {
            path[j][k]=1;
            }
            dfs(start,num);
            cout<<"Case "<<turn<<": The longest path from "<<start<<" has length "<<len[start]<<", finishing at "<<end[start]<<"."<<endl<<endl;
            }
            //system("PAUSE");
            return   0;
            }
            不妨執(zhí)行一下
            5
            3
            1 2
            3 5
            3 1
            2 4
            4 5
            0 0
            先是len[3]=0;end[3]=3;flag[3]=1;
            再執(zhí)行t=dfs(1)+1,
            轉(zhuǎn)入dfs(1);len[1]=0;end[1]=1;flag[1]=1;
            再執(zhí)行t=dfs(2)+1;
            轉(zhuǎn)入dfs(2),len[2]=0;end[2]=2;flag[2]=1;
            再執(zhí)行t=dfs(4)+1
            轉(zhuǎn)入dfs(4),len[4]=0;end[4]=4;flag[4]=1;
            再轉(zhuǎn)入t=dfs(5)+1;
            轉(zhuǎn)入dfs(5),len[5]=0;end[5]=5;flag[5]=1;return(len[5]=0);
            則t=1;t>len[4];len[4]=1;end[4]=end[5]=5;再看4沒了其他相鄰元素。dfs(4)=return(len[4])=1;
            t=dfs(4)+1=2;len[2]=t=2;end[2]=end[4]=5;再看2沒了其他相鄰元素,dfs(2)=return(len(2)=2;
            再看t=dfs(2)+1=3;len[1]=t=3;end[1]=en[2]=5;再看1有沒有其他相鄰元素,dfs(1)=return(len(1)=3
            再執(zhí)行t=dfs(1)+1,len[3]=4;end[3]=end[1]=5;再看3有沒有其他相鄰元素,有dfs(5),已經(jīng)遍歷到了,所以dfs(5)return len【5】。
            沒有影響。
            假設(shè)改為
            5 3 5 2 3 5 3 1 2 4 4 1 0 0 執(zhí)行時會走3->1>這時的1結(jié)點len[1]已經(jīng)求的 3>5>2>4>1len[1]已知了
            posted on 2009-06-29 16:13 luis 閱讀(275) 評論(0)  編輯 收藏 引用 所屬分類: 搜索 、給我啟發(fā)題
            <2012年2月>
            2930311234
            567891011
            12131415161718
            19202122232425
            26272829123
            45678910

            常用鏈接

            留言簿(3)

            隨筆分類

            隨筆檔案

            文章分類

            文章檔案

            友情鏈接

            搜索

            •  

            最新評論

            閱讀排行榜

            評論排行榜

            国产成人综合久久精品红| 国内精品久久久久久久久电影网| 久久久久久亚洲精品成人| 久久99国内精品自在现线| yellow中文字幕久久网| 日韩欧美亚洲国产精品字幕久久久| 久久精品国产免费观看| 99久久精品国产综合一区| 亚洲欧洲精品成人久久曰影片| 亚洲精品白浆高清久久久久久| 日韩亚洲欧美久久久www综合网| 亚洲精品第一综合99久久| 99久久er这里只有精品18| 无码精品久久一区二区三区 | 久久久WWW免费人成精品| 99久久国产精品免费一区二区| 天天爽天天爽天天片a久久网| 亚洲精品国产第一综合99久久| 国产999精品久久久久久| 99久久婷婷国产综合亚洲| 一本色道久久88精品综合| 日韩亚洲国产综合久久久| 狠狠综合久久综合中文88| 国产精品久久永久免费| 久久精品国产99久久无毒不卡| 久久婷婷是五月综合色狠狠| 久久久久亚洲av毛片大| 国产亚洲精午夜久久久久久| 国内精品久久久久影院日本| 久久婷婷激情综合色综合俺也去| 亚洲人成网站999久久久综合| 久久精品综合一区二区三区| 国产精品综合久久第一页| 亚洲一本综合久久| 国产99久久久国产精品~~牛| 久久电影网一区| 亚洲成色999久久网站| 国产精品综合久久第一页| 久久久噜噜噜久久| 久久久久99这里有精品10| 热99RE久久精品这里都是精品免费 |