div1 250pt:

　　题意:100*100的01矩阵，找出来面积最大的“类似国际象棋棋盘”的子矩阵。

　　解法：枚举矩阵宽(水平方向)的起点和终点，然后利用尺取法来找到每个固定宽度下的最大矩阵，不断更新答案。

 // BEGIN CUT HERE

 // END CUT HERE

 #line 5 "TheMatrix.cpp"

 #include<cstdio>

 #include<sstream>

 #include<cstring>

 #include<cstdlib>

 #include<ctime>

 #include<cmath>

 #include<cassert>

 #include<iostream>

 #include<string>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<stack>

 #include<algorithm>

 using namespace std;

 typedef long long ll;

 typedef pair<int,int> pii;

 class TheMatrix

 {

     public:

     bool check(vector<string>& s,int start,int end,int row){

         for(int i = start + ;i <=end;i++)

             if(s[row][i]==s[row][i-])return false;

         return true;

     }

     bool ok(vector<string>& s,int start,int end,int row){

         for(int i=start;i<=end;i++)

             if(s[row][i]==s[row-][i])return false;

         return true;

     }

     int MaxArea(vector <string> s){

     //$CARETPOSITION$

         int answer = ;

         int n=s.size(),m=s[].size();

         for(int i=;i<m;i++)

             for(int j=;j<m;j++){

                 int down=,up=;

                 for(down=;down<n;down++){

                     if(check(s,i,j,down)){

                         up=down+;

                         while(up<n&&ok(s,i,j,up))up++;

                         int area=(j-i+)*(up-down);

                         answer=max(answer,area);

                         down=up-;

                     }

                 }

             }

         return answer;

     }

 // BEGIN CUT HERE

     public:

     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); if ((Case == -) || (Case == )) test_case_6(); if ((Case == -) || (Case == )) test_case_7(); }

     private:

     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }

     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }

     void test_case_0() { string Arr0[] = {"",

  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_1() { string Arr0[] = {""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_2() { string Arr0[] = {""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_3() { string Arr0[] = {"",

  "",

  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_4() { string Arr0[] = {""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_5() { string Arr0[] = {"",

  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_6() { string Arr0[] = {"",

  "",

  "",

  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

     void test_case_7() { string Arr0[] = {"",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  "",

  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, MaxArea(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main(){

     TheMatrix ___test;

     ___test.run_test(-);

     return ;

 }

 // END CUT HERE

250pt

div1 500pt:

　　题意：有个人有个飞机，起初有F升油，有很多个任务，第i个任务消耗duration[i]油，完成之后得到refuel[i]升油，求最多能完成的任务数。

　　解法：先按照refuel降序排列，然后背包。

　　　　why?

　　　　对于两个任务，refuel大的在前面做一定要优于在后面做。假设现在剩余F升油，有两个任务a,b,其中duration[a] > duration[b],如果a先做，那么做完之后会剩余F - duration[a] + refuel[a]升油，如果b先做，会剩余F - duration[b] + refuel[b]升油，也就是说，如果两个都能做完的话，至少要保证F-duration[a]-duraion[b]+refuel[a/b]>=0,

显然先做refuel大的更可能完成。

　　

 // BEGIN CUT HERE

 // END CUT HERE

 #line 5 "AlbertoTheAviator.cpp"

 #include<cstdio>

 #include<cstring>

 #include<cstdlib>

 #include<ctime>

 #include<cmath>

 #include<sstream>

 #include<cassert>

 #include<iostream>

 #include<string>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<stack>

 #include<algorithm>

 using namespace std;

 typedef long long ll;

 typedef pair<int,int> pii;

 int dp[][];

 pii hs[];

 bool cmp(pii a,pii b){

     return a.first > b.first;

 }

 class AlbertoTheAviator

 {

     public:

     int MaximumFlights(int F, vector <int> duration, vector <int> refuel){

     //$CARETPOSITION$

         int n = duration.size();

         for(int i = ;i < n;i++){

             hs[i].first = refuel[i];

             hs[i].second = duration[i];

         }

         sort(hs,hs+n,cmp);

         memset(dp,-,sizeof(dp));

         dp[][F] = ;

         for(int i = ;i < n;i++){

             for(int j=;j<=;j++)dp[i+][j]=dp[i][j];

             for(int j=hs[i].second;j <= ;j++){

                 if(dp[i][j] == -)continue;

                 dp[i+][j-hs[i].second+hs[i].first]=max(dp[i+][j-hs[i].second+hs[i].first],dp[i][j]+);

             }

         }

         int answer = ;

         for(int i = ;i <= ;i++)

             answer = max(answer,dp[n][i]);

         return answer;

     }

 // BEGIN CUT HERE

     public:

     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }

     private:

     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }

     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }

     void test_case_0() { int Arg0 = ; int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

     void test_case_1() { int Arg0 = ; int Arr1[] = {, }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {, }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

     void test_case_2() { int Arg0 = ; int Arr1[] = {, , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {, , , }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

     void test_case_3() { int Arg0 = ; int Arr1[] = {, }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {, }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

     void test_case_4() { int Arg0 = ; int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

     void test_case_5() { int Arg0 = ; int Arr1[] = {, , , , , , , , , , , , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arr2[] = {, , , , , , , , , , , , , }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); int Arg3 = ; verify_case(, Arg3, MaximumFlights(Arg0, Arg1, Arg2)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main(){

     AlbertoTheAviator ___test;

     ___test.run_test(-);

     return ;

 }

 // END CUT HERE

500pt

　　

topcoder srm 610的更多相关文章

  1. topcoder SRM 610 DIV2 TheMatrix

    题目的意思是给一个01的字符串数组,让你去求解满足棋盘条件的最大棋盘 棋盘的条件是: 相邻元素的值不能相同 此题有点像求全1的最大子矩阵,当时求全1的最大子矩阵是用直方图求解的 本题可以利用直方图求解 ...

  2. topcoder SRM 610 DIV2 DivideByZero

    题目的意思是给你一组数,然后不断的进行除法(注意是大数除以小数),然后将得到的结果加入这组数种然后继续进行除法, 直到没有新添加的数为止 此题按照提议模拟即可 注意要保持元素的不同 int Count ...

  3. topcoder srm 610 div2 250

    第一次做tc 的比赛,一点也不懂,虽然题目做出来了, 但是,也没有在比赛的时候提交成功.. 还有,感谢一宁对tc使用的讲解.. 贴一下代码..... #include <cstring> ...

  4. topcoder srm 610 div1

    problem1 link 计算每个格子向上的最大高度.然后每个格子同一行前面的格子以及当前格子作为选取的矩形的最后一行,计算面积并更新答案. problem2 link 对于两个数据$(x_{1}, ...

  5. TopCoder SRM 560 Div 1 - Problem 1000 BoundedOptimization & Codeforces 839 E

    传送门:https://284914869.github.io/AEoj/560.html 题目简述: 定义"项"为两个不同变量相乘. 求一个由多个不同"项"相 ...

  6. Topcoder SRM 643 Div1 250<peter_pan>

    Topcoder SRM 643 Div1 250 Problem 给一个整数N,再给一个vector<long long>v; N可以表示成若干个素数的乘积,N=p0*p1*p2*... ...

  7. Topcoder Srm 726 Div1 Hard

    Topcoder Srm 726 Div1 Hard 解题思路: 问题可以看做一个二分图,左边一个点向右边一段区间连边,匹配了左边一个点就能获得对应的权值,最大化所得到的权值的和. 然后可以证明一个结 ...

  8. TopCoder SRM 667 Div.2题解

    概览: T1 枚举 T2 状压DP T3 DP TopCoder SRM 667 Div.2 T1 解题思路 由于数据范围很小,所以直接枚举所有点,判断是否可行.时间复杂度O(δX × δY),空间复 ...

  9. Topcoder Srm 673 Div2 1000 BearPermutations2

    \(>Topcoder \space Srm \space 673 \space Div2 \space 1000 \space BearPermutations2<\) 题目大意 : 对 ...

随机推荐

  1. MessageBox的使用

    一    函数原型及参数 function MessageBox(hWnd: HWND; Text, Caption: PChar; Type: Word): Integer; hWnd:对话框父窗口 ...

  2. Hibernate-03 关联映射

    学习任务 关联映射 inverse属性.cascade属性 单向的多对一.双向的一对多映射 多对多映射 关联关系 类与类之间最普遍的关系就是关联关系. 单向的关联 双向的关联 单向多对一关联 以Emp ...

  3. mybatis-spring_缓存

    学习之前需要先了解一下什么是mybatis一级缓存? LZ推荐:https://blog.csdn.net/niunai112/article/details/80601793#%E4%B8%80%E ...

  4. pymysql遇到中文编码

    明明数据库里的编码方式和字符集都没有问题,用python插入数据时,数据库里的数据还是乱码的 在数据库中插入时,能够正常显示 那就是python导出的数据存在编码问题,代码如下: # coding: ...

  5. jqury 延迟方法

    $("button").click(function(){    $("#div1").delay("slow").fadeIn();    ...

  6. HDU-1297-Children’s Queue

    Children’s Queue 这道题是排序问题,可以用递归方法解决. 计算F(n): 一:当最后一个是男孩M时候,前面n-1个随便排出来,只要符合规则就可以,即是F(n-1): 二:当最后一个是女 ...

  7. openjudge-4017 爬楼梯

    总时间限制: 1000ms 内存限制: 65536kB 描述 树老师爬楼梯,他可以每次走1级或者2级,输入楼梯的级数,求不同的走法数 例如:楼梯一共有3级,他可以每次都走一级,或者第一次走一级,第二次 ...

  8. Linux文件属性和压缩解压

    目 录 第1章 Linux系统文件的属性    1 1.1 命令ls    1 1.2 inode    2 1.3 文件属性    2 1.4 env命令    2 1.5 注意:    2 第2章 ...

  9. Python面向对象之异常

    异常的概念 程序在运行时,如果python解释器遇见一个错误,就会停止程序的运行,并且提示一些错误信息,这就是异常: 程序遇见错误停止执行并且提示错误信息,这个动作我们称之为抛出(raise)异常: ...

  10. 剑指Offer（书）：重建二叉树

    题目:输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树.假设输入的前序遍历和中序遍历的结果中都不含重复的数字.例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2, ...