Capture the Flag(模拟)

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 1^st team attacks the 2^nd 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 j^th number of the i^th line indicating the j^th team's maintaining status of the i^th 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 L^th team. The relative error or absolute error of the score should be less than 10^-5. 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^-5.

Sample Input

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.

题解：

排序的时候错了下，又开了个结构体，存排名，就过了，意思就是有N个队，每个队Q个防御塔，每个队初始得分是S,C个检查点；

每个检查点输入每个对的得分以及排名；

接下来一个A，A行，每行a,b,c代表a在防御塔c处打了b;

被打的扣N-1分，打的平分这些分；

然后Q行代表防御塔是否完好，坏了要修补，花费N-1，其他好的平分这些分；

还有如果存在多个a,b,c只按一个算；

说的有点啰嗦，就是一个模拟

代码：

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<set>

 #include<vector>

 using namespace std;

 typedef long long LL;

 struct Node{

     double sc;

     int k, rat;

     friend bool operator < (Node a, Node b){

         if(a.sc >= b.sc)return true;

         else return false;

     }

 };

 Node dt[], ans[];

 int vis[][][];

 vector<int>vec[][];

 struct Node1{

     int b, c;

 };

 Node1 vv[];

 int sj[];

 int main(){

     int T, N, Q, S, C;

     scanf("%d", &T);

     while(T--){

         scanf("%d%d%d%d", &N, &Q, &S, &C);

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

             dt[i].k = i, dt[i].sc = S;

         while(C--){

             memset(vis, , sizeof(vis));

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

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

                     vec[i][j].clear();

             int tp = ;

             int A;

             scanf("%d", &A);

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

                 int a, b, c;

                 scanf("%d%d%d", &a, &b, &c);

                 if(!vis[a][b][c]){

                     vis[a][b][c] = ;

                     if(vec[b][c].size() == ){

                         vv[tp].b = b;

                         vv[tp].c = c;

                         tp++;

                     }

                     vec[b][c].push_back(a);

                 }

             }

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

                 dt[vv[j].b].sc -= (N - );

                 for(int i = ; i < vec[vv[j].b][vv[j].c].size(); i++){

                     int a = vec[vv[j].b][vv[j].c][i];

                     dt[a].sc += 1.0*(N - )/vec[vv[j].b][vv[j].c].size();

                 }

             }

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

                 int t = ;

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

                     scanf("%d", sj + i);

                     if(sj[i])t++;

                 }

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

                     if(sj[i] == ){

                         dt[i].sc -= (N - );

                         for(int k = ; k <= N; k++){

                             if(sj[k]){

                                 dt[k].sc += 1.0*(N - )/t;

                             }

                         }

                     }

                 }

             }

             for(int i = ; i <= ; i++)ans[i] = dt[i];

             sort(ans + , ans + N + );

             ans[].rat = ;

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

                 if(fabs(ans[i].sc - ans[i - ].sc) <= 1e-){

                     ans[i].rat = ans[i - ].rat;

                 }

                 else ans[i].rat = i;

             }

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

                 dt[ans[i].k].rat = ans[i].rat;

             }

             int L;

             scanf("%d", &L);

             while(L--){

                 int x;

                 scanf("%d", &x);

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

                     if(dt[i].k == x){

                         printf("%lf %d\n", dt[i].sc, dt[i].rat);

                         break;

                     }

                 }

             }

         }

     }

     return ;

 }