SRM 393(1-250pt)
题意:有m个人投票,每个人在心里对所有候选者排了一个序,比如“210”,则他最想投2号,如果2号已经出局他会投1号,最后投0号,否则弃权不投。选举时进行多轮投票,知道选出winner或者所有人均出局。每轮投票以后,得票最高者所得票数如果严格大于该轮投票人数的50%,则他成为winner,游戏结束;若不大于50%,则将得票最低的人淘汰,如果有多人得票相同且最低,则一起淘汰出局,然后进行下一轮投票。
解法:纯模拟。不过题意很不清晰,有两个地方我都弄了半天猜弄懂。。。就当学习官方题解的代码了。
tag:simulation
// BEGIN CUT HERE
/*
* Author: plum rain
* score :
*/
/* */
// END CUT HERE
#line 11 "InstantRunoffVoting.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack> using namespace std; #define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define mp make_pair
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef pair<int, int> pii;
typedef long long int64; const double eps = 1e-;
const double PI = atan(1.0)*;
const int inf = / ; bool vis[];
int idx[], num[];
pii an[]; bool cmp(pii a, pii b)
{
return a.second < b.second;
} class InstantRunoffVoting
{
public:
int winner(vector <string> v){
clr0 (vis); clr0 (idx);
int n = sz(v), m = sz(v[]);
int cnt = ;
while (cnt < m){
int tmp = n;
clr0 (num);
for (int i = ; i < n; ++ i){
while (idx[i] < m && vis[v[i][idx[i]] - '']) ++ idx[i];
if (idx[i] >= m){
-- tmp; continue;
}
++ num[v[i][idx[i]] - ''];
}
int all = ;
for (int i = ; i < m; ++ i)
if(!vis[i]) an[all++] = mp(i, num[i]);
sort(an, an+all, cmp);
if (an[all-].second * > tmp) return an[all-].first; for (int i = ; i < all; ++ i){
if (an[i].second != an[].second) break;
++ cnt; vis[an[i].first] = ;
}
}
return -;
} // 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() { string Arr0[] = {"","","","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, winner(Arg0)); }
void test_case_1() { string Arr0[] = {"","","","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, winner(Arg0)); }
void test_case_2() { string Arr0[] = {"",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = -; verify_case(, Arg1, winner(Arg0)); }
void test_case_3() { string Arr0[] = {"","",""
,"","",""
,"",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = -; verify_case(, Arg1, winner(Arg0)); }
void test_case_4() { string Arr0[] = {"","","","",""
,"","","","",""
,"","","","",""
,"","","","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, winner(Arg0)); }
void test_case_5() { string Arr0[] = {"","",""
,"","",""
,"","",""
,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, winner(Arg0)); } // END CUT HERE }; // BEGIN CUT HERE
int main()
{
// freopen( "a.out" , "w" , stdout );
InstantRunoffVoting ___test;
___test.run_test(-);
return ;
}
// END CUT HERE
SRM 393(1-250pt)的更多相关文章
- SRM593(1-250pt,500pt)
SRM 593 DIV1 250pt 题意:有如下图所示的平面,每个六边形有坐标.将其中一些六边形染色,要求有边相邻的两个六边形不能染同一种颜色.给定哪些六边形需要染色,问最少需要多少种颜色. 解法: ...
- SRM475 - SRM479(1-250pt,500pt)
SRM 475 DIV1 300pt 题意:玩游戏.给一个棋盘,它有1×n(1行n列,每列标号分别为0,1,2..n-1)的格子,每个格子里面可以放一个棋子,并且给定一个只含三个字母WBR,长度为n的 ...
- SRM468 - SRM469(1-250pt, 500pt)
SRM 468 DIV1 250pt 题意:给出字典,按照一定要求进行查找. 解法:模拟题,暴力即可. tag:water score: 0.... 这是第一次AC的代码: /* * Author: ...
- SRM470 - SRM474(1-250pt,500pt)(471-500pt为最短路,474-500pt未做)
SRM 470 DIV1 250pt 题意:有n个房间排成一排,相邻两个房间之间有一扇关闭着的门(共n-1扇),每个门上都标有‘A’-‘P’的大写字母.给定一个数n,表示第n个房间.有两个人John和 ...
- SRM 609(1-250pt, 1-500pt)
嗯....还是应该坚持写题解的好习惯啊... DIV1 250pt 这难度是回到srm 300+的250了嘛...略 // BEGIN CUT HERE /* * Author: plum rain ...
- SRM 442(1-250pt, 1-500pt)
DIV1 250pt 题意:将一个数表示成质因子相乘的形式,若乘式所含数字的个数为质数,则称A为underprime.比如12 = 2*2*3,则含3个数字,是underprime.求A, B之间un ...
- 记第一次TopCoder, 练习SRM 583 div2 250
今天第一次做topcoder,没有比赛,所以找的最新一期的SRM练习,做了第一道题. 题目大意是说 给一个数字字符串,任意交换两位,使数字变为最小,不能有前导0. 看到题目以后,先想到的找规律,发现要 ...
- SRM 513 2 1000CutTheNumbers(状态压缩)
SRM 513 2 1000CutTheNumbers Problem Statement Manao has a board filled with digits represented as St ...
- SRM 510 2 250TheAlmostLuckyNumbersDivTwo(数位dp)
SRM 510 2 250TheAlmostLuckyNumbersDivTwo Problem Statement John and Brus believe that the digits 4 a ...
随机推荐
- SQL存储过程基于字段名传入的字符串拼接.
--定义存储过程. Create PROCEDURE Usp_Static ), ), --分组字段. ), --统计字段. ), --表头字段. ) --聚会函数. AS ) --存储游标执行的列. ...
- 玩转CSLA.NET小技巧系列一:跳转页面丢失session,如何解决
很少写代码,最近在写代码被登录难倒了,这丫的一直在跟我较劲 每次登录完跳转到首页后还是未登录状态 if (ModelState.IsValid) { bool isSuccess = FI.Finan ...
- c#wiform中KeyDown事件
当首次按下键盘上某个键时发生事件. 例如 private void Form1_KeyDown(object sender, KeyEventArgs e) { if (e.KeyCode == Ke ...
- [转]Delphi执行CMD命令
今天看到有人在问用代码执行CMD命令的问题,就总结一下用法,也算做个备忘. Delphi中,执行命令或者运行一个程序有2个函数,一个是winexec,一个是shellexecute.这两个大家应该都见 ...
- javascript的选项卡
主要用的索引值 首先 写三个按钮 <input type="button" > <input type="button" > <i ...
- mysql如何将一个表导出为excel表格
方法一:进入到mysql的控制台,输入: 1. SELECT * INTO OUTFILE ‘./test.xls‘ FROM tb1 WHERE 1 ORDER BY id DESC LIMIT ...
- asp.net中后台javaScrip的使用
ClientScriptManager csm = Page.ClientScript; //Script标记靠近<form>标签 //csm.Register ...
- SQL Server Analysis Services 数据挖掘(1)
来源: http://technet.microsoft.com/zh-cn/library/dn633476.aspx 假如你有一个购物类的网站,那么你如何给你的客户来推荐产品呢?这个功能在很多 电 ...
- 使用模版引擎填充重复dom元素
引入arttemplate,定义newajax发送跨域请求获得数据,将获得的数据用定义的格式渲染 <!DOCTYPE html><html lang="en"&g ...
- springtest+juint开发测试如下:
项目结构目录如下: UserMapper.java 为接口文件.User 为实体类.UserMapper.xml 为对应mybatis的xml文件.test为对应的测试包 applicationtes ...