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zoj Capture the Flag 比较难的模拟题

来源:程序员人生   发布时间:2015-07-28 08:08:48 阅读次数:3429次

Capture the Flag

Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

In computer security, Capture the Flag (CTF) is a computer security competition. CTF contests are usually designed to serve as an educational exercise to give participants experience in securing a machine, as well as conducting and reacting to the sort of attacks found in the real world. Reverse-engineering, network sniffing, protocol analysis, system administration, programming, and cryptanalysis are all skills which have been required by prior CTF contests at DEF CON. There are two main styles of capture the flag competitions: attack/defense and jeopardy.

In an attack/defense style competition, each team is given a machine (or a small network) to defend on an isolated network. Teams are scored on both their success in defending their assigned machine and on their success in attacking other team's machines. Depending on the nature of the particular CTF game, teams may either be attempting to take an opponent's flag from their machine or teams may be attempting to plant their own flag on their opponent's machine.

Recently, an attack/defense style competition called MCTF held by Marjar University is coming, and there are N teams which participate in the competition. In the beginning, each team has S points as initial score; during the competition, there are some checkpoints which will renew scores for all teams. The rules of the competition are as follows:

  • If a team has been attacked for a service P, they will lose N - 1 points. The lost points will be split equally and be added to the team(s) which attacks successfully. For example, there are 4 teams and Team A has been attacked by Team B and Team C, so Team A will lose 3 points, while Team B and Team C each will get 1.5 points.
  • If a team cannot maintain their service well, they will lose N - 1 points, which will be split equally too and be added to the team(s) which maintains the service well.

The score will be calculated at the checkpoints and then all attacks will be re-calculated. Edward is the organizer of the competition and he needs to write a program to display the scoreboard so the teams can see their scores instantly. But he doesn't know how to write. Please help him!

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains four integers N (2 <= N <= 100) - the number of teams, Q - the number of services (1 <= Q <= 10), S - the initial points (0 <= S <= 100000) and C - the number of checkpoints (1 <= C <= 100).

For each checkpoint, there are several parts:

  • The first line contains an integer A - the number of the successful attacks. Then A lines follow and each line contains a message:
    [The No. of the attacker] [The No. of the defender] [The No. of the service]
    For example, "1 2 3" means the 1st team attacks the 2nd team in service 3 successfully. The No. of teams and services are indexed from 1. You should notice that duplicate messages are invalid because of the rules. Just ignore them.
  • Then there are Q lines and each line contains N integers. The jth number of the ith line indicating the jth team's maintaining status of the ith service, where 1 means well and 0 means not well.
  • Finally there is an integer U (0 <= U <= 100), which describing the number of the queries. The following line contains U integers, which means Edward wants to know the score and the ranking of these teams.

Output

For each query L, output the score and the ranking of the Lth team. The relative error or absolute error of the score should be less than 10. The team with higher score gets higher rank; the teams with the same scores should have the same rank. It is guaranteed that the scores of any two teams are either the same or with a difference greater than 10.

Sample Input

1 4 2 2500 5 0 1 1 1 1 1 1 1 1 4 1 2 3 4 2 1 2 1 3 2 1 1 1 1 1 1 1 1 1 4 1 2 3 4 1 1 2 2 1 1 1 1 1 1 1 0 4 1 2 3 4 0 0 0 0 0 0 0 0 0 4 1 2 3 4 0 1 1 1 1 1 1 1 1 2 1 4

Sample Output

2500.00000000 1 2500.00000000 1 2500.00000000 1 2500.00000000 1 2501.50000000 1 2497.00000000 4 2501.50000000 1 2500.00000000 3 2505.50000000 1 2495.00000000 4 2502.50000000 2 2497.00000000 3 2499.50000000 1 2489.00000000 4 2496.50000000 2 2491.00000000 3 2499.50000000 1 2491.00000000 3

Hint

For C++ users, kindly use scanf to avoid TLE for huge inputs.


链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5527


题意:

n个队伍,q台服务器,s为初始分数,5次操作


每次操作中有A次攻击

每次攻击 a b ser,表示 a队攻击 b队,通过服务器ser。疏忽重复出现 一样的 a b ser。

攻击完后,计算分数要把每台服务器上的攻击分开计算。被攻击的队伍-(n⑴)分,把这n⑴分 平分给攻击他的队伍。


然后是q行n列

表示 每台服务器上,n支队伍有无 保护好 各自领域。  把没保护好的 队伍扣n⑴分,把扣的总分平分给 保护好的队伍。


然后1个U,询问U个队伍的 分数和排名,然后输出。  排序的时候注意精度, 如 0.123451和0.123452 实际上不相等,但是精度为1e⑸时,算是相等的


#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> #include <malloc.h> #include <ctype.h> #include <math.h> #include <string> #include <iostream> #include <algorithm> using namespace std; #include <stack> #include <queue> #include <vector> #include <deque> #include <set> #include <map> using namespace std; #define maxn 505 #define eps 1e⑸ struct nodedef { double point; int id; int rank; }; nodedef nod[110]; int ok[11][110]; int vis[110][110][15]; vector <int> vec[11][110]; int has[110]; int cmp1(nodedef a,nodedef b) { return a.point>b.point; } int cmp2(nodedef a,nodedef b) { return a.id<b.id; } int main() { int t; scanf("%d",&t); while(t--) { int n,q,s,c; scanf("%d%d%d%d",&n,&q,&s,&c); for(int i=1;i<=n;i++) { nod[i].point=s; nod[i].id=i; } while(c--) { int aa; scanf("%d",&aa); memset(vis,0,sizeof vis); for(int i=1;i<=q;i++) { for(int j=1;j<=n;j++) { vec[i][j].clear(); } } for(int i=0;i<aa;i++)//A次服务器攻击,各个服务器之间是没有相互影响的,所以ser要开1个维度,然后把攻击b的所有队伍存在在1起 { int a,b,ser; scanf("%d %d %d",&a,&b,&ser); if(vis[a][b][ser]==0) //重复的命令去掉 vec[ser][b].push_back(a); vis[a][b][ser]=1; } for(int i=1;i<=q;i++) { for(int j=1;j<=n;j++) { if(vec[i][j].size()) { nod[j].point-=(n⑴); for(int k=0;k<vec[i][j].size();k++)//依照规则,攻击的人平分这n⑴ 分 { nod[vec[i][j][k]].point+=1.0*(n⑴)/vec[i][j].size(); } } } } for(int i=1;i<=q;i++)//不同服务器直接 不相互影响 { for(int j=1;j<=n;j++) { int tem; scanf("%d",&ok[i][j]); } int li=0; int yi=0; for(int j=1;j<=n;j++) { if(ok[i][j]) yi++; else { nod[j].point-=(n⑴); li++; } } for(int j=1;j<=n;j++) { if(ok[i][j]) nod[j].point+=1.0*(n⑴)*li/yi; //把li个被破坏的服务器数*n⑴分 平分被yi个保护好的服务器 } } sort(nod+1,nod+1+n,cmp1); nod[1].rank=1; for(int i=2;i<=n;i++) { if((nod[i⑴].point-nod[i].point)<eps)//要注意的地方,由于精度问题,所以这里要<eps //如 0.123451和0.123452 实际上不相等,但是精度为1e⑸时,算是相等的 nod[i].rank=nod[i⑴].rank; else nod[i].rank=i; } int cha; scanf("%d",&cha); sort(nod+1,nod+1+n,cmp2); for(int i=1;i<=cha;i++) { int id; scanf("%d",&id); printf("%f %d ",nod[id].point,nod[id].rank); } } } return 0; }






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