1056. Mice and Rice (25)
2015-12-06 11:11
309 查看
使用队列来模拟还在的人
时间限制30 ms
内存限制65536 kB
代码长度限制16000 B
判题程序Standard作者CHEN, Yue
Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.
First the playing order is randomly decided for NP programmers. Then every NG programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every NGwinners are then grouped in the next match until a final winner is determined.
For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: NP and NG (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG mice at the end of the player's list, then all the mice left will be put into the last group. The second line contains NP distinct non-negative numbers Wi (i=0,...NP-1) where each Wiis the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,...NP-1 (assume that the programmers are numbered from 0 to NP-1). All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.
Sample Input:
来源: <http://www.patest.cn/contests/pat-a-practise/1056>
[/code]
来自为知笔记(Wiz)
时间限制30 ms
内存限制65536 kB
代码长度限制16000 B
判题程序Standard作者CHEN, Yue
Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.
First the playing order is randomly decided for NP programmers. Then every NG programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every NGwinners are then grouped in the next match until a final winner is determined.
For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: NP and NG (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG mice at the end of the player's list, then all the mice left will be put into the last group. The second line contains NP distinct non-negative numbers Wi (i=0,...NP-1) where each Wiis the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,...NP-1 (assume that the programmers are numbered from 0 to NP-1). All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.
Sample Input:
11 3 25 18 0 46 37 3 19 22 57 56 10 6 0 8 7 10 5 9 1 4 2 3Sample Output:
5 5 5 2 5 5 5 3 1 3 5
来源: <http://www.patest.cn/contests/pat-a-practise/1056>
#include<iostream>
#include<stdio.h>
#include<vector>
#pragma warning(disable:4996)
#include<queue>
using namespace std;
int np, ng;
int weight[1010] = { 0 };
int rank1[1010] = { 0 };
int mark[1010] = { 0 };
queue<int> order, nextRound;
int main(void) {
freopen("Text.txt", "r", stdin);
cin >> np >> ng;
for (int i = 0; i < np; i++) {
cin >> weight[i];
}
for (int i = 0; i < np; i++) {
int temp;
cin >> temp;
order.push(temp);
}
while (true)
{
int nextRoundCnt = order.size() / ng;
if (order.size() %ng != 0)
nextRoundCnt++;
int outRank = nextRoundCnt + 1;
for (int i = 0; i < nextRoundCnt; i++) {
int max = -1, maxIndex = -1;
for (int j = 0; j < ng &&order.empty()==false; j++) {
if (weight[order.front()] > max) {
if (maxIndex != -1) {
rank1[maxIndex] = outRank;
}
max = weight[order.front()];
maxIndex = order.front();
}
else {
rank1[order.front()] = outRank;
}
order.pop();
}
nextRound.push(maxIndex);
}
int size = nextRound.size();
for (int i = 0; i < size; i++) {
order.push(nextRound.front());
nextRound.pop();
}
if (order.size() == 1) {
rank1[order.front()] = 1;
break;
}
}
for (int i = 0; i < np; i++) {
cout << rank1[i];
if (i != np - 1)
cout << " ";
}
return 0;
}
[/code]
来自为知笔记(Wiz)
相关文章推荐
- 报错apachectl -t
- java内部类实例化
- 1055. The World's Richest (25)
- ubuntu开机自动启动程序设置
- 蓝桥杯 错误票据(排序)
- 高效能人士的七个习惯
- Mac之Git/GitHub使用(3)——Fork A Repo
- jquery入门
- Android FM学习中的模块 FM启动过程
- C++子类在成员函数中不要转型为父类
- iconfont代替png图片
- 堆排序
- curl命令详解
- 优先队列 两个堆的维护
- 理解Golang包导入
- 1054. The Dominant Color (20)
- win7不能正常启动,只能进入安全模式
- nc命令使用详解
- 1053. Path of Equal Weight (30)
- 工作中用到的小算法,计算两日期间隔xx年xx月xx天