山东省第七届省赛 D题:Swiss-system tournament(归并排序)
Description
A Swiss-system tournament is a tournament which uses a non-elimination format. The first tournament of this type was a chess tournament in Zurich in 1895, hence the name "Swiss system". The tournament will be held based on following rules.
2*N contestants (indexed 1, 2, ..., 2*N) will have R rounds matches. Before the first round, every contestant has an origin score. After every match, winner will get 1 score and loser will get 0 score. Before and after every round, contestants will be sorted by their scores in descending order. Two contestants with the same score will be sorted by their index with ascending order.
In every round, contestants will have match based on the sorted list. The first place versus the second place, the third place versus the forth place, ..., the Kth place versus the (K + 1)th place, ..., the (2*N - 1)th place versus (2*N)th place.
Now given the origin score and the ability of every contestant, we want to know the index of the Qth place contestant. We ensured that there won’t be two contestants with the same ability and the contestant with higher ability will always win the match.
Input
Multiple test cases. The first line contains a positive integer T (T<=10) indicating the number of test cases.
For each test case, the first line contains three positive integers N (N <= 100,000), R (R <= 50), Q (Q <= 2*N), separated by space.
The second line contains 2*N non-negative integers, s1, s2, ..., s2*N, si (si<= 108) indicates the origin score of constant indexed i.
The third line contains 2*N positive integers, a1, a2, ..., a2*N, ai (ai<= 108) indicates the ability of constant indexed i.
Output
One line per case, an integer indicates the index of the Qth place contestant after R round matches.
Sample Input
1 2 4 2 7 6 6 7 10 5 20 15
Sample Output
1
题意:
给出2*n个人,每次第1个人和第2个人比赛,第3个人和第4个人比赛…进行r轮比赛,每轮比完赛都进行排名,求出最后第q名是谁。给出每个人的积分 s[i],每个人的能力a[i],能力大的获胜,获胜的加1分,输了的不加分。
题解:
此题利用归并的思想,每轮比赛将赢者放一组,输者放一组,然后进行合并。
代码:
#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cstdlib>
using namespace std;
#define is_lower(c) (c>='a' && c<='z')
#define is_upper(c) (c>='A' && c<='Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>='0' && c<='9')
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\
cin.tie();\
cout.tie();
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
;
const ll inf_ll=(ll)1e18;
const ll mod=1000000007LL;
;
;
struct node {
int pos;
int ability;
int score;
}q[*N+],Q[*N+];
bool cmp(node a,node b)
{
if(a.score == b.score)
return a.pos<b.pos;
return a.score>b.score;
}
int n,r,p;
int main()
{
int T;
cin>>T;
while(T--)
{
,r1 = ;
scanf("%d%d%d",&n,&r,&p);
; i <= n * ; i++)
{
scanf("%d",&q[i].score);
q[i].pos = i;
}
; i <= n * ; i++)
scanf("%d",&q[i].ability);
sort(q+,q++*n,cmp);
; j <= r; j++){
l1 = ;r1 = n;
; i <= * n; i += )
{
].ability) {
q[i].score++;
Q[++l1] = q[i];
Q[++r1] = q[i+];
}else {
q[i+].score++;
Q[++l1] = q[i+];
Q[++r1] = q[i];
}
}
;
,rr=n+;
while(ll<=l1&&rr<=r1){
if(Q[ll].score == Q[rr].score)
{
if(Q[ll].pos<Q[rr].pos)
q[++flag] = Q[ll++];
else
q[++flag] = Q[rr++];
}else {
if(Q[ll].score>Q[rr].score)
q[++flag] = Q[ll++];
else
q[++flag] = Q[rr++];
}
}
while(ll<=l1)
q[++flag] = Q[ll++];
while(rr<=r1)
q[++flag] = Q[rr++];
}
printf("%d\n",q[p].pos);
}
;
}
山东省第七届省赛 D题:Swiss-system tournament(归并排序)的更多相关文章
- 山东省第七届ACM竞赛 J题 Execution of Paladin (题意啊)
题意:鱼人是炉石里的一支强大种族,在探险者协会里,圣骑士有了一张新牌,叫亡者归来,效果是召唤本轮游戏中7个已死鱼人.如果死掉的不足7个,那么召唤的数量就会不足7. 鱼人有很多,下面的4个是: 寒光智者 ...
- Sdut 2164 Binomial Coeffcients (组合数学) (山东省ACM第二届省赛 D 题)
Binomial Coeffcients TimeLimit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 输入 输出 示例输入 1 1 10 2 9 ...
- Sdut 2165 Crack Mathmen(数论)(山东省ACM第二届省赛E 题)
Crack Mathmen TimeLimit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Since mathmen take security ...
- ACM Sdut 2158 Hello World!(数学题,排序) (山东省ACM第一届省赛C题)
题目描述 We know thatIvan gives Saya three problems to solve (Problem F), and this is the firstproblem. ...
- 山东省第七届ACM竞赛 C题 Proxy (Dijkstra算法,单源路径最短问题)
题意:给定0-n+1个点,和m条边,让你找到一条从0到n+1的最短路,输出与0相连的结点... 析:很明显么,是Dijkstra算法,不过特殊的是要输出与0相连的边,所以我们倒着搜,也是从n+1找到0 ...
- 山东省第六届省赛 H题:Square Number
Description In mathematics, a square number is an integer that is the square of an integer. In other ...
- 山东省第七届ACM省赛------Memory Leak
Memory Leak Time Limit: 2000MS Memory limit: 131072K 题目描述 Memory Leak is a well-known kind of bug in ...
- 山东省第七届ACM省赛------Reversed Words
Reversed Words Time Limit: 2000MS Memory limit: 131072K 题目描述 Some aliens are learning English. They ...
- 山东省第七届ACM省赛------Triple Nim
Triple Nim Time Limit: 2000MS Memory limit: 65536K 题目描述 Alice and Bob are always playing all kinds o ...
随机推荐
- Java中WeakHashMap实现原理深究
一.前言 我发现Java很多开源框架都使用了WeakHashMap,刚开始没怎么去注意,只知道它里面存储的值会随时间的推移慢慢减少(在 WeakHashMap 中,当某个“弱键”不再正常使用时,会被从 ...
- doget,doPost在底层走的是service
doget,doPost在底层走的是service 因为在源码上 先执行service方法 然后再调用doget,doPost方法
- Citrix Netscaler负载均衡算法
Citrix Netscaler负载均衡算法 http://blog.51cto.com/caojin/1926308 众所周知,作为新一代应用交付产品的Citrix Netscaler具有业内领先的 ...
- [洛谷P4390][BOI2007]Mokia 摩基亚
题目大意: 维护一个W*W的矩阵,每次操作可以增加某格子的权值,或询问某子矩阵的总权值. 题解:CDQ分治,把询问拆成四个小矩形 卡点:无 C++ Code: #include <cstdio& ...
- Tourists——圆方树
CF487E Tourists 一般图,带修求所有简单路径代价. 简单路径,不能经过同一个点两次,那么每个V-DCC出去就不能再回来了. 所以可以圆方树,然后方点维护一下V-DCC内的最小值. 那么, ...
- ng websocket
ng使用websocket 1.安装依赖库npm install ws --save 2.安装类型定义文件 npm install @types/ws --save 3.编写服务 import { I ...
- codeforces 1077F1
题目:https://codeforces.com/contest/1077/problem/F1 题意: 你有n幅画,第i幅画的好看程度为ai,再给你两个数字k,x 表示你要从中选出刚好x幅画,并且 ...
- 模拟实现jdk动态代理
实现步骤 1.生成代理类的源代码 2.将源代码保存到磁盘 3.使用JavaCompiler编译源代码生成.class字节码文件 4.使用JavaCompiler编译源代码生成.class字节码文件 5 ...
- 【数据结构】bzoj3747Kinoman
Description 共有m部电影,编号为1~m,第i部电影的好看值为w[i]. 在n天之中(从1~n编号)每天会放映一部电影,第i天放映的是第f[i]部. 你可以选择l,r(1<=l< ...
- noip2016 提高组
T1 玩具谜题 题目传送门 这道题直接模拟就好了哇 233 #include<cstdio> #include<cstring> #include<algorithm&g ...