ZOJ 3757 Alice and Bob and Cue Sports（模拟题）

A - Alice and Bob and Cue Sports

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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Appoint description: System Crawler  (2014-03-02)

Description

Alice and Bob both love playing games. Now, Alice and Bob are playing a cue sport like Nine-ball Pool. Two players move in turn. A move is that one player use the cue ball to hit the target ball for pocketing the target ball. At the beginning of game, there are one cue ball and n object balls. Each object ball has a number (a positive integer) on it, and no two object balls have the same number. In a move, the target ball is a object ball still on the table with the smallest number. If the player does the following things in a move, a foul is called and a penalty point is added to the opposite's point:

• Cue ball do not hit any object ball. Penalty: the number of target ball.
• Cue ball is not pocketed and hit at least one ball, but do not hit target ball first or hit more than one object ball first at the same time. Penalty: the largest number of ball in the first hit balls.
• Cue ball is pocketed and hit at least one ball. Penalty: the largest number of ball in the first hit balls.

If the player pockets the target ball without a foul, a point will be added to the player's points, which is the sum of numbers of the ball pocketed in this move, and the player can make another move. But, if the player pockets the target ball with a foul or pockets at least one object ball not including the target ball, this point will be added to the opposite's points. If cue ball is pocketed, it will be pull out of the pocket and put on the table again while object balls will not be put on the table again after they are pocketed, even if the player pocketed them with a foul.

Now given n object balls and m moves, Bob wants to write a program to calculate the final points of Alice and him after these m moves. Because of Lady's first Rule, Alice always moves first.

Input

There are multiple cases. The first line of each case consists of two integers, n and m (1 ≤ n ≤ 1000, 0 ≤ m ≤ 1000), as described above. The second line consists of n positive integers. Each integer a_i means the number of i^th object ball (1 ≤ a_i ≤ 10000). There are 2m lines followed. The (2i-1)^th and (2i)^th lines describe the i^th move. The (2i-1)^th line first includes an integer p, meaning cue ball hits p balls first. There are p integers followed in the same line, describing the hit balls. The (2i)^th line first includes an integer q, meaning there are q balls pocketed. There are q integers followed in the same line, describing the pocketed balls (0 means cue ball). It's guaranteed that there is no impossible data.

Output

For each case, output one line as the format: "AP : BP", where AP means Alice's points and BP means Bob's points.

Sample Input

3 32 3 51 21 21 31 51 31 3

Sample Output

2 : 8

题意：打桌球，按规则罚分，尼玛啊，模拟题就算了吧，还要是题意题，犯规之后除了罚那些分，还要罚全部进球的分啊，你妹啊

题解：就是一个大模拟题啊。。。尼玛啊，浪费时间啊，题意原来都没懂啊，可以不坑我吗啊，看别人代码才知道看错题意啊

（仅以此文纪念浪费的时间和逝去的码力和死去的脑细胞啊）

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1008],b[1008],c[1008],mark[1008],res[2];int main(){    int n,m,i,j,k,now,x,y,temp,flag,flag2;    while(~scanf("%d%d",&n,&m))    {        for(i=0; i<n; i++) scanf("%d",a+i);        sort(a,a+n);        res[0]=res[1]=0;        for(now=k=0; k<m; k++)        {            scanf("%d",&x);            for(i=0; i<x; i++) scanf("%d",b+i);            scanf("%d",&y);            for(i=0; i<y; i++) scanf("%d",c+i);            flag2=0;            if(!x)            {                res[!now]+=a[0];            }            else            {                for(flag=i=0; i<y; i++) if(c[i]==0)                    {                        flag=1;                        break;                    }                if(flag)                {                    for(temp=i=0; i<x; i++) if(b[i]>temp) temp=b[i];                    res[!now]+=temp;                }                else                {                    if(x>1)                    {                        for(temp=i=0; i<x; i++) if(b[i]>temp) temp=b[i];                        res[!now]+=temp;                    }                    else                    {                        if(b[0]!=a[0])                        {                            for(temp=i=0; i<x; i++) if(b[i]>temp) temp=b[i];                            res[!now]+=temp;                        }                        else                        {                            for(i=0; i<y; i++) if(c[i]==a[0])                                {                                    flag2=1;                                    break;                                }                        }                    }                }            }            if(flag2)            {                for(i=0; i<y; i++) res[now]+=c[i];            }            else            {                now=!now;                for(i=0; i<y; i++) res[now]+=c[i];            }            memset(mark,0,sizeof(mark));            for(i=0; i<y; i++)            {                if(c[i]==0) continue;                for(j=0; j<n; j++) if(a[j]==c[i])                    {                        mark[j]=1;                        break;                    }            }            for(i=j=0; i<n; i++)            {                if(mark[i]) continue;                a[j++]=a[i];            }            n=j;        }        printf("%d : %d\n",res[0],res[1]);    }    return 0;}