• <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>

            oyjpArt ACM/ICPC算法程序設(shè)計(jì)空間

            // I am new in programming, welcome to my blog
            I am oyjpart(alpc12, 四城)
            posts - 224, comments - 694, trackbacks - 0, articles - 6

            PKU 1011 Sticks

            Posted on 2006-11-30 00:34 oyjpart 閱讀(4020) 評論(15)  編輯 收藏 引用 所屬分類: ACM/ICPC或其他比賽
            這道題作的我真的是悲喜交加阿。。。做完之后。。。長舒一口氣。。推薦大家去做!!!

            Sticks
            Time Limit:1000MS? Memory Limit:10000K
            Total Submit:18973 Accepted:4421

            Description
            George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.

            Input
            The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

            Output
            The output should contains the smallest possible length of original sticks, one per line.

            Sample Input

            9
            5 2 1 5 2 1 5 2 1
            4
            1 2 3 4
            0
            

            Sample Output

            6
            5 















            1。我從大到小搜索了哇 沒用。。。
            2。我想 用預(yù)先得到所有可拼湊長度來HASH 發(fā)現(xiàn)太大...
            3。然后我想對每個(gè)長棍分開搜索...
            4。后來我又用記錄數(shù)目的方法搜 似乎更慢...
            終于發(fā)現(xiàn)真正重要的剪枝!
            1.當(dāng)一個(gè)正好可以填滿的時(shí)候 就不用考慮比他小的去填了
            2.一大段的第一個(gè)小段如果不成立直接返回到上一大段
            這才是重要的剪枝
            同時(shí)還有一個(gè) 要預(yù)防反復(fù)搜索同一關(guān)鍵碼 給出下面的測試數(shù)據(jù)
            64
            40 40 40 40 40 40 40 40 40 40 40 40 40
            40 40 40 40 40 40 40 40 40 40 43 42 42
            41 10 4 40 40 40 40 40 40 40 40 40 40
            40 40 40 40 40 40 40 40 40 40 40 40 40
            40 40 40 40 40 40 40 40 40 40 40 40
            0
            呵呵 其實(shí)AC的程序里面有一大部分都過不了這個(gè)數(shù)據(jù)!包括0MSAC的!

            呵呵 過了之后 心情好啊~`哈哈
            //Solution
            //by optimistic
            #include <stdio.h>
            #include <stdlib.h>
            #include <string.h>
            int nss;
            int ss[70];
            int used[70];
            int totss;
            int maxss;
            int len;
            int cmp(const void * a, const void * b)
            {
            ?return (*(int *)b) - (*(int *)a);
            }
            int search(int times, int rest, int pos)
            {
            ?int flag = 0;
            ?if(rest == len) flag = 1; //第一種剪枝
            ?int i;
            ?if(times == totss/len) return 1;
            ?for(i = pos; i<nss; i++)
            ??if(!used[i])
            ??{
            ???if(rest == ss[i])
            ???{
            ????used[i] = 1;
            ????if(search(times+1, len, 0))?return 1;
            ????used[i] = 0;
            ??? ?return 0;????????????????????? //第二種剪枝???????????????????????????????????????????????????????????????????
            ???}
            ???else if(ss[i]<rest)
            ???{
            ????used[i] = 1;
            ????if(search(times, rest-ss[i], i+1)) return 1;
            ????used[i] = 0;
            ????if(flag) return 0;
            ????while(ss[i] == ss[i+1]) i++;
            ???}
            ???else if(flag) return 0;
            ??}
            ?return 0;
            }
            int main()
            {
            //?freopen("t.in", "r", stdin);
            ?int i;
            ?while(scanf("%d", &nss), nss>0)
            ?{
            ??memset(ss, 0, sizeof(ss));
            ??totss = maxss = 0;
            ??for(i=0; i<nss; i++)
            ??{
            ???scanf("%d", &ss[i]);
            ???totss += ss[i];
            ???if(ss[i]>maxss) maxss = ss[i];
            ??}
            ??qsort(ss, 70, sizeof(ss[0]), cmp);
            ??for(i=maxss; i<=totss; i++)
            ??{
            ???if(i==totss)
            ???{printf("%d\n", totss); break;}
            ???if(totss%i==0)
            ???{?????
            ????memset(used, 0, sizeof(used));
            ????len = i;

            ????if(search(1, len, 0)) {printf("%d\n", i); break;}
            ???}
            ??}
            ?}
            ?return 0;
            }




            Feedback

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2007-04-13 23:40 by Jun Wang
            if(search(1, len, 0)) {printf("%d\n", i); break;}
            是不是要改成 if(search(0, len, 0)) {printf("%d\n", i); break;} ??

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2007-07-22 19:28 by Typhoooooooooooooooooooooooooooooooooon
            感謝你那兩個(gè)重要的剪枝哈

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2007-10-24 11:00 by delguoqing
            你上面這個(gè)測試數(shù)據(jù)的ouput是多少?

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-04 23:46 by mango
            你測這個(gè)... 你的半天也出不來...
            64
            40 40 30 35 35 26 15 40 40 40 40 40 40 40 40 40 40 40 40 40 40

            40 40 43 42 42 41 10 4 40 40 40 40 40 40 40 40 40 40 40 40 40

            40 25 39 46 40 10 4 40 40 37 18 17 16 15 40 40 40 40 40 40 40

            40

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-05 09:02 by oyjpart
            的確啊,很強(qiáng)大的數(shù)據(jù)啊

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-05 18:42 by zoyi
            答案是454~~可是我的程序居然是wa~5555555555

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-05 20:10 by oyjpart
            哦?你怎么知道答案啊

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-06 19:43 by zoyi
            我的程序跑出來的啊~~難道你懷疑我跑的是錯(cuò)誤的???哈哈@oyjpart

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-07 12:49 by mango
            哎......這個(gè)數(shù)據(jù)真變態(tài)...煩死了 呵呵

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-07 21:06 by oyjpart
            哦。。。你過題了沒

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-05-22 15:59 by zsong
            我跑出454了,很快,不過也是wa

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-08-12 20:45 by zjh777007
            誰能告訴我
            “一大段的第一個(gè)小段如果不成立直接返回到上一大段”
            什么意思?

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-08-12 20:53 by oyjpart
            現(xiàn)在由一個(gè)當(dāng)前情況S.
            這個(gè)時(shí)候比如有一個(gè)大棍子,長度為20,現(xiàn)在嘗試在其中放入一個(gè)長度為5的小棍子。結(jié)果深搜得到的結(jié)果是不可行,則認(rèn)為當(dāng)前情況是無解的。
            因?yàn)檫@個(gè)5長度的小棍子放不了這個(gè)大棍子,絕對放不了任何一根大棍子。

            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-11-30 11:29 by abc
            #include<iostream>
            #include<string>
            #include<fstream>
            #include<vector>
            #include<ctime>
            using namespace std;
            fstream fin("e:\\office\\txt\\acmin.txt");
            int l[65];
            int s,t,min,sum=0,len;
            int max2;
            int count=0;
            int c=0;
            static int used[65];
            int search(int leni, int s){
            if(leni==0) return 100;
            t=s;
            int count=0;
            while(l[t]>leni||used[t]){
            t--;
            if(t==0){
            //break;
            return 0;
            }
            }
            used[t]=1;
            count++;
            if(l[t]<=leni){
            int se=search(leni-l[t],s-1);

            if(se>=100){
            count+=(se-100);
            //if((leni-l[t])!=0){
            //if(count<s){
            c=1;
            for(int i=1;i<=s;i++)c*=used[i];
            if(!c){
            int sea=search(len,s-1);
            if(sea>=100){
            count+=(sea-100);
            return count+100;
            }/*else{
            return 0;
            }*/
            //}
            }
            }
            /*else{
            return 0;
            }*/
            }
            count+=100;
            return count;
            }
            int main(){
            cin>>s;
            while(s){

            max2=0;
            sum=0;
            count=0;
            for(int i=0;i<65;i++)used[i]=0;
            for(int i=1;i<=s;i++){
            cin>>l[i];
            if(l[i]>max2){
            max2=l[i];
            }
            sum+=l[i];
            }
            for(int i=0;i<=s;i++){
            for(int j=0;j<s-i;j++){
            if(l[j]>l[j+1]){
            int temp=l[j];
            l[j]=l[j+1];
            l[j+1]=temp;
            }
            }
            }
            cout<<"sum"<<sum<<endl;
            for(int i=max2;i<=sum;i++){
            len=i;

            for(int j=0;j<65;j++)used[j]=0;
            if(sum%i!=0)continue;

            int k=search(i,s);

            if(k>100){

            if((k-100)==s){cout<<i<<endl;break;}
            }else{
            continue;
            }
            }
            cin>>s;
            }

            }













            # re: PKU 1011 Sticks   回復(fù)  更多評論   

            2008-11-30 11:30 by abc
            各位高手幫忙看一下上面的程序有什么錯(cuò)誤,萬分感激!
            色妞色综合久久夜夜| 性高朝久久久久久久久久| 久久午夜无码鲁丝片秋霞| 久久亚洲精品无码aⅴ大香 | 国产精品女同久久久久电影院| 久久久网中文字幕| 热久久这里只有精品| 久久精品无码专区免费青青| 久久久久久久久久久久久久| 亚洲欧洲中文日韩久久AV乱码| 99热都是精品久久久久久| 国产成人久久AV免费| 亚洲国产高清精品线久久| 久久ww精品w免费人成| 性做久久久久久久久| 久久91精品国产91久久户| 久久国产一区二区| 亚洲国产另类久久久精品| 久久久精品2019免费观看| 亚洲国产成人精品女人久久久 | 久久久久国产精品人妻| 久久九九青青国产精品| 日韩av无码久久精品免费| 精品国产乱码久久久久久1区2区 | 久久精品国产99久久久| 香蕉久久夜色精品国产2020| 欧美精品九九99久久在观看| 99久久精品九九亚洲精品| 性高朝久久久久久久久久| 国产精品欧美久久久久无广告| 久久99国产精品一区二区| 无码AV波多野结衣久久| 精品伊人久久大线蕉色首页| 人人狠狠综合久久亚洲高清| 久久AⅤ人妻少妇嫩草影院| 午夜人妻久久久久久久久| 日韩人妻无码一区二区三区久久99 | 久久久久国产精品| 青草影院天堂男人久久| www性久久久com| 精品综合久久久久久97超人|