Capture the Flag ZOJ - 3879（模擬題）
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 Tindicating 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 10-5. 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
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
題意：首先給你四個數，n，q，s，c，代表著隊伍的個數，伺服器的個數，最初的分數，以及檢查點的個數；
然後以下c行:
對於每一個檢查點c：
1、首先有一個A，代表有A次攻擊，對於每次攻擊，有三個數a，b，c，a攻擊了c伺服器上的b；
對於每個伺服器上的隊伍，受攻擊的會減少（n-1）分，並且會平均分配給攻擊這個隊伍的那些隊伍；
還有一點就是重複的訊息無效。（這點很坑）。
2、然後有q行01串，一行n位，代表每個伺服器上的每個隊伍的狀態，如果為0，就代表處於失分狀態，就會失去（n-1）分，然後平均分配給這一行上狀態是1的那些隊伍
3、最後有k個數字，為詢問的隊伍的個數，然後k個數字，代表要你輸出這k個人的分數以及排名，分數相同的排名也一樣。
總之題目比較長比較醜，不過理解了之後，還是一個比較水的模擬題；
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const double eps=1e-5; 8 int t,n,q,c,u; 9 double s; 10 bool vis[105][105][15]; 11 int a,x,y,z,ser[105][15],book[105],cnt,p[105]; 12 struct team{ 13 int i; 14 double s; 15 }te[105]; 16 17 inline bool cmp1(team a,team b){ 18 return a.s>b.s; 19 } 20 21 inline bool cmp2(team a,team b){ 22 return a.i<b.i; 23 } 24 25 26 int main(){ 27 scanf("%d",&t); 28 while(t--){ 29 scanf("%d%d%lf%d",&n,&q,&s,&c); 30 for(int i=1;i<=n;i++) te[i].s=s,te[i].i=i; 31 while(c--){ 32 scanf("%d",&a); 33 for(int i=1;i<=n;++i){ 34 for(int j=1;j<=n;++j){ 35 for(int k=1;k<=q;++k) 36 { 37 vis[i][j][k]=0; 38 ser[j][k]=0; 39 } 40 } 41 } 42 for(int i=1;i<=a;++i){ 43 scanf("%d%d%d",&x,&y,&z); 44 if(!vis[x][y][z]) 45 { 46 vis[x][y][z]=1; 47 ser[y][z]++; 48 } 49 50 } 51 for(int j=1;j<=n;++j) 52 for(int k=1;k<=q;++k) 53 if(ser[j][k]) 54 { 55 te[j].s-=(n-1); 56 for(int i=1;i<=n;++i) 57 if(vis[i][j][k]) 58 te[i].s+=(n-1)/(double)ser[j][k]; 59 } 60 for(int j=1;j<=q;++j) 61 { 62 cnt=0; 63 for(int i=1;i<=n;++i) 64 { 65 scanf("%d",&book[i]); 66 if(!book[i]){ 67 cnt++; 68 te[i].s-=(n-1); 69 } 70 } 71 for(int i=1;i<=n;++i) 72 if(book[i]) te[i].s+=(double)(n-1)*cnt/(n-cnt); 73 } 74 sort(te+1,te+n+1,cmp1); 75 for(int i=1;i<=n;++i) 76 { 77 if(i==1||fabs(te[i].s-te[i-1].s)>eps) 78 p[te[i].i]=i; 79 else 80 p[te[i].i]=p[te[i-1].i]; 81 } 82 sort(te+1,te+n+1,cmp2); 83 scanf("%d",&u); 84 while(u--){ 85 int temp; 86 scanf("%d",&temp); 87 printf("%.8f %d\n",te[temp].s,p[temp]); 88 } 89 } 90 } 91 return 0; 92 }