每年省赛必有的一道模拟题，描述都是非常的长，题目都是蛮好写的...
sigh... 比赛的时候没有写出这道题目 :(
题意：首先输入4个数，n,q,p,c代表有n个队伍，q个服务器，每支队伍的初始分数p，
还有c次操作对于每次操作，首先输入一个k，代表k次攻击每次攻击有三个数，a,b,c，
代表a通过c服务器成功的攻击了b对于每次成功的攻击，b会失去n-1分，
这n-1分会平分给通过同一服务器攻击b的几支队伍然后是q行，每行n个数，
分别代表1~n是否成功维护了，1为成功，0为不成功不成功的队伍会失去n-1分，
这失去的分会平分给成功维护的那些队伍然后输入k个数，询问这k支队伍的分数
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0) using namespace std; typedef long long ll ;
typedef unsigned long long ull ;
typedef unsigned int uint ;
typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e- ;
const int N = ;
const int M = * ;
const ll P = 10000000097ll ;
const int MAXN = ; struct node {
int id, rank;
double score;
} a[]; int n, q, c, t;
double p;
bool vis[][][], hsh[]; int cmp1 (node a,node b) {
return a.score > b.score;
} int cmp2 (node a,node b) {
return a.id < b.id;
} int main () {
int i, j, k;
scanf ("%d",&t);
while (t--) {
scanf ("%d%d%lf%d", &n, &q, &p, &c);
for (i = ; i <= n; ++i) {
a[i].id = i;
a[i].rank = ;
a[i].score = p;
}
while (c--) {
scanf("%d",&k);
memset (vis, , sizeof (vis));
while(k--) {
int atk, def, sev;
scanf("%d%d%d", &atk, &def, &sev);
if (vis[atk][def][sev]) continue;
vis[atk][def][sev] = true;
}
for (i = ; i <= q; ++i) {
for (j = ; j <= n; ++j) { //防守方
int cnt = ;
for (k = ; k <= n; ++k) { //攻击方
if (vis[k][j][i]) //统计攻击j的队伍有几支
++cnt;
}
if (!cnt) continue;
double ss = 1.0 * (n - ) / cnt;//平分
a[j].score -= (n - ); //防守方失去n-1
for (k = ; k <= n; ++k) {
if (vis[k][j][i]) //攻击方得到分数
a[k].score += ss;
}
}
} for (i = ; i <= q; ++i) { //服务器
memset (hsh, , sizeof (hsh));
int cnt = ;
for (j = ; j <= n; ++j) {
int x;
scanf ("%d",&x);
if (x) { //成功维护的队伍数
hsh[j] = true;
++cnt;
} else {
hsh[j] = false;
a[j].score -= (n - );
}
}
if(cnt == n) continue;
double ss = 1.0 * (n - ) / cnt;
ss = ss * (n - cnt);
for (j = ; j <= n; ++j) {
if (hsh[j]) {
a[j].score += ss;
}
}
}
sort (a + , a + n + , cmp1);
for (i = ; i <= n; ++i) { //更新排名
if (i != ) {
if (fabs(a[i].score - a[i - ].score) < 1e-)
a[i].rank = a[i - ].rank;
else
a[i].rank = i;
} else {
a[i].rank = i;
}
}
sort (a + , a + n + , cmp2);
scanf ("%d", &k);
while (k--) {
int x;
scanf ("%d",&x);
printf ("%.5f %d\n",a[x].score, a[x].rank);
}
}
} return ;
}