• <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>

            ACM___________________________

            ______________白白の屋
            posts - 182, comments - 102, trackbacks - 0, articles - 0
            <2010年10月>
            262728293012
            3456789
            10111213141516
            17181920212223
            24252627282930
            31123456

            常用鏈接

            留言簿(24)

            隨筆分類(332)

            隨筆檔案(182)

            FRIENDS

            搜索

            積分與排名

            最新隨筆

            最新評論

            閱讀排行榜

            評論排行榜

            MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋    

             

             

             

            題目地址 :

                  http://acm.hdu.edu.cn/showproblem.php?pid=1698

            題目描述 : 

            Just a Hook

            Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
            Total Submission(s): 3841    Accepted Submission(s): 1675


            Problem Description
            In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.



            Now Pudge wants to do some operations on the hook.

            Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
            The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

            For each cupreous stick, the value is 1.
            For each silver stick, the value is 2.
            For each golden stick, the value is 3.

            Pudge wants to know the total value of the hook after performing the operations.
            You may consider the original hook is made up of cupreous sticks.
             

            Input
            The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
            For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
            Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.
             

            Output
            For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.
             

            Sample Input
            1 10 2 1 5 2 5 9 3
             

            Sample Output
            Case 1: The total value of the hook is 24.
             

             

            標準的線段樹,  成段更新 ,......     具體看 代碼 注釋 .

             

            代碼如下 :

             /*

            Coded By  : MiYu

            Link      : http://www.cnblogs.com/MiYu  || http://m.shnenglu.com/MiYu

            Author By : MiYu

            Test      : 1

            Program   : 1698

            */

            //#pragma warning( disable:4789 )

            #include <iostream>

            #include <algorithm>

            #include <string>

            #include <set>

            #include <map>

            #include <utility>

            #include <queue>

            #include <stack>

            #include <list>

            #include <vector>

            #include <cstdio>

            #include <cstdlib>

            #include <cstring>

            #include <cmath>

            using namespace std;


            typedef struct seg_tree{

            int left, right, col;

            bool cov; //標記當前線段是否被覆蓋, 如果true, 表示這一段線段的值都為 col. false則相反

            int mid (){ return (left + right) >> 1; }

            }SEG;

            SEG seg[300010];

            void creat ( int beg, int end, int rt = 1 ){

            seg[rt].left = beg;

            seg[rt].right = end;

            seg[rt].col =  1;

            seg[rt].cov =  true;

            if ( beg == end ) return;

            int mid = seg[rt].mid();

            creat ( beg, mid, rt << 1 );

            creat ( mid + 1, end, ( rt << 1 ) + 1 );

            }

            void modify ( int beg, int end, int val, int rt = 1 ){

            int LL = rt << 1;

            int RR = ( rt << 1 ) + 1;

            if ( seg[rt].left == beg && seg[rt].right == end ){ //線段被覆蓋, 標記 cov 為true  

            seg[rt].cov = true;

            seg[rt].col = val;

            return ;

            }

            if ( seg[rt].cov ){ //如果線段曾經被覆蓋,  標記 false, 將col往下傳  

            seg[rt].cov = false;

            seg[LL].col = seg[RR].col = seg[rt].col;

            seg[LL].cov = seg[RR].cov = true;

            }

            int mid = seg[rt].mid();

            if ( end <= mid ){

            modify ( beg, end, val, LL );

            } else if ( beg > mid ) {

            modify ( beg, end, val, RR );

            } else {

            modify ( beg, mid, val, LL );

            modify ( mid + 1, end, val, RR );

            }

            }

            int quy ( int beg, int end, int rt = 1 ){

            if ( seg[rt].cov ){  // 線段如果是被覆蓋的 , 直接返回這一段區間的值

            return ( seg[rt].right - seg[rt].left + 1 ) * seg[rt].col;

            }

            int mid = seg[rt].mid();

            return quy ( beg, mid, rt << 1 ) + quy ( mid + 1, end, ( rt << 1 ) + 1 );

            }


            int main ()

            {

            int T, ca = 1;

            scanf ( "%d", &T );

            while ( T -- ){

            int N;

            scanf ( "%d", &N );

            creat ( 1, N );

            int M;

            scanf ( "%d", &M );

            for ( int i = 1; i <= M; ++ i ){

            int beg, end, val;

            scanf ( "%d%d%d", &beg, &end, &val );

            modify ( beg, end, val );

            }

            printf ( "Case %d: The total value of the hook is %d.\n", ca++,quy( 1, N ) );

            }

                return 0;

            }


            /*

            1

            10

            2

            1 5 2

            5 9 3

            */



            /*    此為一牛人代碼 , 速度 非常快 !!!! 0rz.........

            #include<stdio.h>

            int a[100001][3],c[100001];

            int main()

            {

                int t,i,j,n,m,sum,v,w=1;

                scanf("%d",&t);

                while(t--&&scanf("%d %d",&n,&m))

                {

                    sum=0;

                    for(i=1;i<=m;i++)

                    scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]);

                    for(i=1;i<=n;i++)

                    {

                        v=1;

                        for(j=m;j>=1;j--)

                        {

                            if(a[j][0]<=i&&a[j][1]>=i)

                            {

                                v=a[j][2];

                                break;

                            }

                        }

                        sum+=v;

                    }

                    printf("Case %d: The total value of the hook is %d.\n",w++,sum);

                }

            }



            */

             

            Feedback

            # re: HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU[未登錄]  回復  更多評論   

            2010-09-18 11:18 by bb
            下面的代碼只是剛好數據不能卡把?復雜度O(n*m)~

            # re: HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU  回復  更多評論   

            2010-09-18 11:42 by MiYu
            Accepted 1698 437MS 4300K 2117 B C++
            這是 線段樹 的 AC 判定,
            Accepted 1698 218MS 1360K 630 B C++
            這是后面方法的 AC 判定, 快了 一倍

            # re: HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU[未登錄]  回復  更多評論   

            2010-09-18 17:39 by bb
            不是呀,只是OJ數據沒卡到這方法~~

            # re: HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU  回復  更多評論   

            2010-10-30 07:56 by MiYu
            我覺得 哪方法 很牛B =. =
            久久久久久久久无码精品亚洲日韩| 狠狠色丁香婷婷久久综合不卡| 亚洲精品综合久久| 亚洲av伊人久久综合密臀性色| 久久99精品国产自在现线小黄鸭| 久久精品视频网| 99久久99久久精品国产片果冻| AV无码久久久久不卡网站下载| 久久夜色撩人精品国产| 国产麻豆精品久久一二三| 亚洲国产成人精品91久久久| 九九久久99综合一区二区| 久久人妻AV中文字幕| 欧美久久亚洲精品| 99久久国产综合精品成人影院| 国产亚洲精久久久久久无码77777| 99久久成人18免费网站| 狼狼综合久久久久综合网| 亚洲日韩欧美一区久久久久我| 蜜桃麻豆www久久| 精品久久久久久无码中文字幕一区| 国产精品成人久久久| 久久精品国产欧美日韩| 久久国产乱子精品免费女| 久久精品国产亚洲AV大全| 亚洲综合精品香蕉久久网| 国内精品伊人久久久久妇| 最新久久免费视频| 久久国产精品无| 久久久久这里只有精品| 久久av免费天堂小草播放| 99久久www免费人成精品| 久久精品免费观看| 国产亚洲美女精品久久久| 精品久久久无码中文字幕天天| 久久成人18免费网站| 久久久中文字幕日本| 久久精品国产国产精品四凭| 久久久WWW免费人成精品| 一本久久综合亚洲鲁鲁五月天亚洲欧美一区二区 | 久久无码精品一区二区三区|