UVALive 5025 Arranging Your Team

Arranging Your Team

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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Description

Your country has qualified for the FIFA 2010 South Africa World Cup. As the coach, you have made up the 23 men squad. Now you must select 11 of them as the starters. As is well known, there are four positions in soccer: goalkeeper, defender, midfielder and striker. Your favorite formation is 4-4-2, that is, you should choose 4 defenders, 4 midfielders, 2 strikers, and of course, 1 goalkeeper. As a retired ACMer, you want to write a program to help you make decision. Each person's ability has been evaluated as a positive integer. And what's more, for some special pairs of persons, if the two people are both on the field, there will be an additional effect (positive or negative). Now you should choose the 11 persons to make the total value maximum.

Input

There are multiple test cases, separated by an empty line. The first 23 lines of each test case indicate each person's name S_i, ability valueV_i, and position. The length of each name is no more than 30, and there are no whitespaces in the names. All the names are different. The ability values are positive integers and no more than 100. The position is one of ``goalkeeper", ``defender", ``midfielder" and ``striker".

Then an integer M indicates that there are M special pairs. Each of the following M lines contains S_i, S_j and C_ij, means that if S_i and S_j are both on the field, the additional profit is C_ij. (- 100C_ij100). S_i and S_j are different strings, and must be in the previous 23 names. All the (S_i, S_j) pairs are different.

Output

Output one line for each test case, indicating the maximum total ability values, that is, the total ability values of the 11 persons plus the additional effects. If you cannot choose a 4-4-2 formation, output ``impossible" instead.

Sample Input

Buffon 90 goalkeeper De_Sanctis 80 goalkeeper Marchetti 80 goalkeeper Zambrotta 90 defender Cannavaro 90 defender Chiellini 90 defender Maggio 90 defender Bonucci 80 defender Criscito 80 defender Bocchetti 80 defender Pirlo 90 midfielderGattuso 90 midfielder De_Rossi 90 midfielder Montolivo 90 midfielder Camoranesi 80 midfielderPalombo 80 midfielder Marchisio 80 midfielder Pepe 80 midfielder Iaquinta 90 striker Di_Natale 90 striker Gilardino 80 striker Quagliarella 80 striker Pazzini 80 striker 1 Pirlo Quagliarella 50 ZhangSan01 50 goalkeeperZhangSan02 50 defender ZhangSan03 50 defender ZhangSan04 50 defender ZhangSan05 50 defender ZhangSan06 50 defender ZhangSan07 50 defender ZhangSan08 50 defender ZhangSan09 50 defender ZhangSan10 50 defender ZhangSan11 50 defender ZhangSan12 50 defender ZhangSan13 50 defender ZhangSan14 50 defender ZhangSan15 50 defender ZhangSan16 50 midfielderZhangSan17 50 midfielderZhangSan18 50 midfielderZhangSan19 50 midfielderZhangSan20 50 midfielderZhangSan21 50 midfielderZhangSan22 50 midfielderZhangSan23 50 midfielder0

Sample Output

1030 impossible

题解及代码：

/*参考了一下别人的代码，才写出来的，感觉他的代码很优美，简洁清爽的感觉。附上网址：http://alwa.name/blog/?p=47*/#include <iostream>#include <cstdio>#include <map>#include <cstring>#include <string>using namespace std;int ans=0;map<string,int>M;map<string,int>N;int vis[30];int add[25][25];struct node{    int d;    int en;}z[25];void solve(int src,int de,int mi,int st,int go){    if(de==4&&mi==4&&st==2&&go==1)    {        int sum=0;        for(int i=0;i<23;i++)        if(vis[i])        {            sum+=z[i].d;            for(int j=0;j<23;j++)            if(vis[j])            {                sum+=add[i][j];            }        }        if(ans<sum) ans=sum;        return;    }    for(int i=src;i<23;i++)    if(!vis[i])    {        if(z[i].en==1&&de==4) continue;        if(z[i].en==2&&mi==4) continue;        if(z[i].en==3&&st==2) continue;        if(z[i].en==4&&go==1) continue;        vis[i]=1;        if(z[i].en==1) solve(i+1,de+1,mi,st,go);        if(z[i].en==2) solve(i+1,de,mi+1,st,go);        if(z[i].en==3) solve(i+1,de,mi,st+1,go);        if(z[i].en==4) solve(i+1,de,mi,st,go+1);        vis[i]=0;    }}int main(){    string s="defender";    N[s]=1;    s="midfielder";    N[s]=2;    s="striker";    N[s]=3;    s="goalkeeper";    N[s]=4;    string l,r;    int d,n;    while(cin>>l>>d>>r)    {        M.clear();        M[l]=0;        z[0].d=d;        z[0].en=N[r];        for(int i=1;i<23;i++)        {            cin>>l>>d>>r;            M[l]=i;            z[i].d=d;            z[i].en=N[r];        }        int num_de=0,num_mi=0,num_st=0,num_go=0;        for(int i=0;i<23;i++)        {            if(z[i].en==1) num_de++;            if(z[i].en==2) num_mi++;            if(z[i].en==3) num_st++;            if(z[i].en==4) num_go++;        }        scanf("%d",&n);        memset(add,0,sizeof(add));        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            cin>>l>>r>>d;            add[M[l]][M[r]]=d;        }        if(num_de<4||num_mi<4||num_st<2||num_go<1)        {            printf("impossible\n");            continue;        }        ans=-100000;//初始化为负数，最后ans可能为负数，wa了一次        solve(0,0,0,0,0);        printf("%d\n",ans);    }    return 0;}