Comments for Greenmoon55's Blog http://greenmoon55.com Just another WordPress weblog Mon, 14 May 2012 11:09:12 +0000 hourly 1 http://wordpress.org/?v=3.3.2 Comment on 火场逃生记 by Greenmoon55 http://greenmoon55.com/escaped-from-a-fire/comment-page-1/#comment-5049 Greenmoon55 Mon, 14 May 2012 11:09:12 +0000 http://greenmoon55.com/?p=216#comment-5049 你从随机文章里看到的吧... 好像当时很淡定，消防车走了我又回来玩游戏了... 你从随机文章里看到的吧…
好像当时很淡定，消防车走了我又回来玩游戏了…

]]> Comment on 火场逃生记 by EUYUIL http://greenmoon55.com/escaped-from-a-fire/comment-page-1/#comment-5027 EUYUIL Sun, 13 May 2012 13:42:20 +0000 http://greenmoon55.com/?p=216#comment-5027 惊险啊。我从来没有经历过火灾…… 惊险啊。我从来没有经历过火灾……

]]> Comment on Google Code Jam 2012 by EUYUIL http://greenmoon55.com/google-code-jam-2012/comment-page-1/#comment-5026 EUYUIL Sun, 13 May 2012 13:40:05 +0000 http://greenmoon55.com/?p=366#comment-5026 我无脑地参加了第一场，跪了。去金华也没时间参加后两场…… 我无脑地参加了第一场，跪了。去金华也没时间参加后两场……

]]> Comment on HDU3682 To Be an Dream Architect by ACM/ICPC 2010 杭州现场赛 To Be a Dream Architect 题解 ‹ EUYUIL http://greenmoon55.com/hdu3682-to-be-an-dream-architect/comment-page-1/#comment-4950 ACM/ICPC 2010 杭州现场赛 To Be a Dream Architect 题解 ‹ EUYUIL Thu, 10 May 2012 07:43:50 +0000 http://greenmoon55.com/?p=338#comment-4950 [...] jQuery(document).ready(function($){ if (navigator.platform == "iPad") return; jQuery("img").not(".cycle img").lazyload({ effect:"fadeIn", placeholder: "http://euyuil.com/wp-content/plugins/jquery-image-lazy-loading/images/grey.gif" }); }); MathJax.Hub.Config({ extensions: ["tex2jax.js"], jax: ["input/TeX", "output/HTML-CSS"], tex2jax: { inlineMath: [ ['$','$'], ["\(","\)"], ["[latex]","[/latex]"] ], displayMath: [ ['$$','$$'], ["\[","\]"] ], processEscapes: true }, "HTML-CSS": { availableFonts: ["TeX"] } }); LiuyueICPC SolutionsSoftware OpusGuestbookSearchAboutACM/ICPC 2010 杭州现场赛 To Be a Dream Architect 题解2012-02-15 21:17:18 • ICPC, 区域赛, 现场赛, 空间几何, 题解 • 已有 2 条评论 google_ad_client = "ca-pub-7808863410544596"; /* EUYUIL.COM 文章内容中 */ google_ad_slot = "7087084322"; google_ad_width = 300; google_ad_height = 250; 这道题的题意是给出一个 N × N 的正方体，这个正方体是由 1 × 1 的小正方体垒成的。然后，输入数据会给出若干个命令，这些命令都表示一条平行于某坐标轴的直线，命令的效果是让这个直线上的所有方块消失掉。最后，问消失了多少小方块。似乎这道题原题标题犯了个小小的语法错误……关于这道题的解法，Greenmoon55 说，可以用 vector + sort + unique 系列的方法解决这个问题，还提到 set 可能会超时超内存[GRN'11]。我想原因是插入删除过程其实是 O(logn) 的，并且使用动态内存分配频繁，各种冗余数据比较多吧。这种方法的时间复杂度是 O(n2logn).我的解法是另一种，我使用一个数组 planes[3][N][N], 记录输入的每一条直线。在每执行一条直线的命令的时候，判断某一个方块是否已经被其它直线去掉了。所以这就降了一维，复杂度是 O(n2). 但是现场赛时，我可能会采取 Greenmoon55 的做法。源代码123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657#include <cstdio> #include <cstring> using namespace std;const int N = 1001;bool planes[3][N][N];int n, m;inline bool is_coord_valid(int x, int y){    if (x >= 1 && x <= n && y >= 0 && y <= n)        return true;    return false;}inline int count_cubes(int i0, int x, int y){    int i1 = (i0 + 1) % 3, i2 = (i0 + 2) % 3;    if (planes[i0][x][y])        return 0; // Already counted.     planes[i0][x][y] = true;    int result = 0;    for (int z = 1; z <= n; ++z)        if (!planes[i1][y][z] && !planes[i2][z][x])            ++result;    return result;}int main(){    int T, x, y, r; char a, b, buf[32];    scanf("%d", &T);    while (T--)    {        scanf("%d %d", &n, &m); r = 0;        memset(planes, false, sizeof planes);        for (int i = 0; i < m; ++i)        {            gets(buf); // Skip EOLN.             scanf("%c=%d,%c=%d", &a, &x, &b, &y);            if (a == 'X' && b == 'Y') r += count_cubes(0, x, y);            else if (a == 'Y' && b == 'X') r += count_cubes(0, y, x);            else if (a == 'Y' && b == 'Z') r += count_cubes(1, x, y);            else if (a == 'Z' && b == 'Y') r += count_cubes(1, y, x);            else if (a == 'Z' && b == 'X') r += count_cubes(2, x, y);            else if (a == 'X' && b == 'Z') r += count_cubes(2, y, x);        }        printf("%dn", r);    }    return 0;}参考[GRN'11] Greenmoon55:《HDU3682 To Be an Dream Architect》原创文章，转载请注明来源：http://euyuil.com/3250/acm-icpc-2010-hangzhou-to-be-a-dream-architect/ 搜索框载入中…… google.load('search', '1', {language : 'zh-CN'}); google.setOnLoadCallback(function() { var customSearchControl = new google.search.CustomSearchControl('001309449942216446108:hg0oww9xcq0'); customSearchControl.setResultSetSize(google.search.Search.FILTERED_CSE_RESULTSET); var options = new google.search.DrawOptions(); options.setAutoComplete(true); options.enableSearchboxOnly("http://euyuil.com/search/", "keyword"); customSearchControl.draw('cse-search-form', options); }, true); [...] [...] jQuery(document).ready(function($){ if (navigator.platform == "iPad") return; jQuery("img").not(".cycle img").lazyload({ effect:"fadeIn", placeholder: "http://euyuil.com/wp-content/plugins/jquery-image-lazy-loading/images/grey.gif&quot; }); }); MathJax.Hub.Config({ extensions: ["tex2jax.js"], jax: ["input/TeX", "output/HTML-CSS"], tex2jax: { inlineMath: [ ['$','$'], ["\(","\)"], ["[latex]","[/latex]"] ], displayMath: [ ['$$','$$'], ["\[","\]"] ], processEscapes: true }, "HTML-CSS": { availableFonts: ["TeX"] } }); LiuyueICPC SolutionsSoftware OpusGuestbookSearchAboutACM/ICPC 2010 杭州现场赛 To Be a Dream Architect 题解2012-02-15 21:17:18 • ICPC, 区域赛, 现场赛, 空间几何, 题解 • 已有 2 条评论 google_ad_client = "ca-pub-7808863410544596"; /* EUYUIL.COM 文章内容中 */ google_ad_slot = "7087084322"; google_ad_width = 300; google_ad_height = 250; 这道题的题意是给出一个 N × N 的正方体，这个正方体是由 1 × 1 的小正方体垒成的。然后，输入数据会给出若干个命令，这些命令都表示一条平行于某坐标轴的直线，命令的效果是让这个直线上的所有方块消失掉。最后，问消失了多少小方块。似乎这道题原题标题犯了个小小的语法错误……关于这道题的解法，Greenmoon55 说，可以用 vector + sort + unique 系列的方法解决这个问题，还提到 set 可能会超时超内存[GRN'11]。我想原因是插入删除过程其实是 O(logn) 的，并且使用动态内存分配频繁，各种冗余数据比较多吧。这种方法的时间复杂度是 O(n2logn).我的解法是另一种，我使用一个数组 planes[3][N][N], 记录输入的每一条直线。在每执行一条直线的命令的时候，判断某一个方块是否已经被其它直线去掉了。所以这就降了一维，复杂度是 O(n2). 但是现场赛时，我可能会采取 Greenmoon55 的做法。源代码123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657#include <cstdio> #include <cstring> using namespace std;const int N = 1001;bool planes[3][N][N];int n, m;inline bool is_coord_valid(int x, int y){    if (x >= 1 && x <= n && y >= 0 && y <= n)        return true;    return false;}inline int count_cubes(int i0, int x, int y){    int i1 = (i0 + 1) % 3, i2 = (i0 + 2) % 3;    if (planes[i0][x][y])        return 0; // Already counted.     planes[i0][x][y] = true;    int result = 0;    for (int z = 1; z <= n; ++z)        if (!planes[i1][y][z] && !planes[i2][z][x])            ++result;    return result;}int main(){    int T, x, y, r; char a, b, buf[32];    scanf("%d", &T);    while (T–)    {        scanf("%d %d", &n, &m); r = 0;        memset(planes, false, sizeof planes);        for (int i = 0; i < m; ++i)        {            gets(buf); // Skip EOLN.             scanf("%c=%d,%c=%d", &a, &x, &b, &y);            if (a == 'X' && b == 'Y') r += count_cubes(0, x, y);            else if (a == 'Y' && b == 'X') r += count_cubes(0, y, x);            else if (a == 'Y' && b == 'Z') r += count_cubes(1, x, y);            else if (a == 'Z' && b == 'Y') r += count_cubes(1, y, x);            else if (a == 'Z' && b == 'X') r += count_cubes(2, x, y);            else if (a == 'X' && b == 'Z') r += count_cubes(2, y, x);        }        printf("%dn", r);    }    return 0;}参考[GRN'11] Greenmoon55:《HDU3682 To Be an Dream Architect》原创文章，转载请注明来源：http://euyuil.com/3250/acm-icpc-2010-hangzhou-to-be-a-dream-architect/ 搜索框载入中…… google.load('search', '1', {language : 'zh-CN'}); google.setOnLoadCallback(function() { var customSearchControl = new google.search.CustomSearchControl('001309449942216446108:hg0oww9xcq0'); customSearchControl.setResultSetSize(google.search.Search.FILTERED_CSE_RESULTSET); var options = new google.search.DrawOptions(); options.setAutoComplete(true); options.enableSearchboxOnly("http://euyuil.com/search/&quot;, "keyword"); customSearchControl.draw('cse-search-form', options); }, true); [...]

]]> Comment on HDU3682 To Be an Dream Architect by EUYUIL http://greenmoon55.com/hdu3682-to-be-an-dream-architect/comment-page-1/#comment-4338 EUYUIL Wed, 15 Feb 2012 06:41:24 +0000 http://greenmoon55.com/?p=338#comment-4338 这种解法真牛逼。 这种解法真牛逼。

]]> Comment on HDU3687 National Day Parade by ACM/ICPC 2010 杭州现场赛 National Day Parade 题解 ‹ EUYUIL http://greenmoon55.com/hdu3687-national-day-parade/comment-page-1/#comment-4337 ACM/ICPC 2010 杭州现场赛 National Day Parade 题解 ‹ EUYUIL Wed, 15 Feb 2012 06:32:03 +0000 http://greenmoon55.com/?p=339#comment-4337 [...] jQuery(document).ready(function($){ if (navigator.platform == "iPad") return; jQuery("img").not(".cycle img").lazyload({ effect:"fadeIn", placeholder: "http://euyuil.com/wp-content/plugins/jquery-image-lazy-loading/images/grey.gif" }); }); MathJax.Hub.Config({ extensions: ["tex2jax.js"], jax: ["input/TeX", "output/HTML-CSS"], tex2jax: { inlineMath: [ ['$','$'], ["\(","\)"], ["[latex]","[/latex]"] ], displayMath: [ ['$$','$$'], ["\[","\]"] ], processEscapes: true }, "HTML-CSS": { availableFonts: ["TeX"] } }); LiuyueACM SolutionsSoftware OpusGuestbookSearchAboutACM/ICPC 2010 杭州现场赛 National Day Parade 题解2012-02-15 14:44:19 • ACM, 区域赛, 水题, 现场赛, 题解 • 抢沙发这是在 Greenmoon55′s Blog 挖掘出来的水题一则。果断凑数。思路是把每行的学生的列坐标分别排序，枚举左边一列的列号即可。代码如下：#crayon-4f3b544841637 .crayon-plain { font-size: 12px !important; line-height: 16px !important;}123456789101112131415161718192021222324252627282930313233343536373839404142434445464748#include <cmath> #include <cstdio> #include <climits> #include <cstring> #include <iostream> #include <algorithm> using namespace std;const int N = 222;int i2j[N][N], cnt[N], n, m, n2;int main(){    // freopen("input.txt", "r", stdin);     while (cin >> n >> m)    {        if (n == 0 && m == 0)            break;        memset(cnt, 0, sizeof cnt);        n2 = n * n;        for (int i = 0; i < n2; ++i)        {            int k, a; cin >> k; --k;            cin >> a; --a; i2j[k][cnt[k]++] = a;        }        for (int i = 0; i < n; ++i)            sort(i2j[i], i2j[i] + n);        int rmost = m - n, sum, result = INT_MAX;        for (int i = 0; i <= rmost; ++i)        {            sum = 0;            for (int j = 0; j < n; ++j)                for (int k = 0; k < n; ++k)                    sum += abs(i2j[j][k] - (i + k));            result = min(result, sum);        }        cout << result << endl;    }    return 0;}原创文章，转载请注明来源：http://euyuil.com/3245/acm-icpc-2010-hangzhou-national-day-parade/ 搜索框载入中…… google.load('search', '1', {language : 'zh-CN'}); google.setOnLoadCallback(function() { var customSearchControl = new google.search.CustomSearchControl('001309449942216446108:hg0oww9xcq0'); customSearchControl.setResultSetSize(google.search.Search.FILTERED_CSE_RESULTSET); var options = new google.search.DrawOptions(); options.setAutoComplete(true); options.enableSearchboxOnly("http://euyuil.com/search/", "keyword"); customSearchControl.draw('cse-search-form', options); }, true); [...] [...] jQuery(document).ready(function($){ if (navigator.platform == "iPad") return; jQuery("img").not(".cycle img").lazyload({ effect:"fadeIn", placeholder: "http://euyuil.com/wp-content/plugins/jquery-image-lazy-loading/images/grey.gif&quot; }); }); MathJax.Hub.Config({ extensions: ["tex2jax.js"], jax: ["input/TeX", "output/HTML-CSS"], tex2jax: { inlineMath: [ ['$','$'], ["\(","\)"], ["[latex]","[/latex]"] ], displayMath: [ ['$$','$$'], ["\[","\]"] ], processEscapes: true }, "HTML-CSS": { availableFonts: ["TeX"] } }); LiuyueACM SolutionsSoftware OpusGuestbookSearchAboutACM/ICPC 2010 杭州现场赛 National Day Parade 题解2012-02-15 14:44:19 • ACM, 区域赛, 水题, 现场赛, 题解 • 抢沙发这是在 Greenmoon55′s Blog 挖掘出来的水题一则。果断凑数。思路是把每行的学生的列坐标分别排序，枚举左边一列的列号即可。代码如下：#crayon-4f3b544841637 .crayon-plain { font-size: 12px !important; line-height: 16px !important;}123456789101112131415161718192021222324252627282930313233343536373839404142434445464748#include <cmath> #include <cstdio> #include <climits> #include <cstring> #include <iostream> #include <algorithm> using namespace std;const int N = 222;int i2j[N][N], cnt[N], n, m, n2;int main(){    // freopen("input.txt", "r", stdin);     while (cin >> n >> m)    {        if (n == 0 && m == 0)            break;        memset(cnt, 0, sizeof cnt);        n2 = n * n;        for (int i = 0; i < n2; ++i)        {            int k, a; cin >> k; –k;            cin >> a; –a; i2j[k][cnt[k]++] = a;        }        for (int i = 0; i < n; ++i)            sort(i2j[i], i2j[i] + n);        int rmost = m – n, sum, result = INT_MAX;        for (int i = 0; i <= rmost; ++i)        {            sum = 0;            for (int j = 0; j < n; ++j)                for (int k = 0; k < n; ++k)                    sum += abs(i2j[j][k] – (i + k));            result = min(result, sum);        }        cout << result << endl;    }    return 0;}原创文章，转载请注明来源：http://euyuil.com/3245/acm-icpc-2010-hangzhou-national-day-parade/ 搜索框载入中…… google.load('search', '1', {language : 'zh-CN'}); google.setOnLoadCallback(function() { var customSearchControl = new google.search.CustomSearchControl('001309449942216446108:hg0oww9xcq0'); customSearchControl.setResultSetSize(google.search.Search.FILTERED_CSE_RESULTSET); var options = new google.search.DrawOptions(); options.setAutoComplete(true); options.enableSearchboxOnly("http://euyuil.com/search/&quot;, "keyword"); customSearchControl.draw('cse-search-form', options); }, true); [...]

]]> Comment on Guestbook by 纳米黑客 http://greenmoon55.com/guestbook/comment-page-1/#comment-4327 纳米黑客 Mon, 13 Feb 2012 03:50:44 +0000 http://greenmoon55.com/?page_id=23#comment-4327 神牛来换链接 神牛来换链接

]]> Comment on HDU4121 Xiangqi by EUYUIL http://greenmoon55.com/hdu4121-xiangqi/comment-page-1/#comment-4282 EUYUIL Mon, 06 Feb 2012 08:57:58 +0000 http://greenmoon55.com/?p=332#comment-4282 共勉啊~！ 共勉啊~！

]]> Comment on Guestbook by 包子剑客 http://greenmoon55.com/guestbook/comment-page-1/#comment-4279 包子剑客 Sun, 05 Feb 2012 15:05:27 +0000 http://greenmoon55.com/?page_id=23#comment-4279 看我邮箱就知道了。。。。。。 这验证又来了。。。。。 看我邮箱就知道了。。。。。。

这验证又来了。。。。。

]]> Comment on Guestbook by Greenmoon55 http://greenmoon55.com/guestbook/comment-page-1/#comment-4278 Greenmoon55 Sun, 05 Feb 2012 15:04:14 +0000 http://greenmoon55.com/?page_id=23#comment-4278 这个名字好熟悉呀～在好多地方见过好多次。。 嗯，确实应该考虑换验证码了= = 这个名字好熟悉呀～在好多地方见过好多次。。
嗯，确实应该考虑换验证码了= =

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