hdu3720 Arranging Your Team
Arranging Your Team
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1218 Accepted Submission(s): 360
Then an integer M indicates that there are M special pairs. Each of the following M lines contains Si, Sj and Cij, means that if Si and Sj are both on the field, the additional profit is Cij. (-100 ≤ Cij ≤ 100). Si and Sj are different strings, and must be in the previous 23 names. All the (Si, Sj) pairs
are different.
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 midfielder
Gattuso 90 midfielder
De_Rossi 90 midfielder
Montolivo 90 midfielder
Camoranesi 80 midfielder
Palombo 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 goalkeeper
ZhangSan02 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 midfielder
ZhangSan17 50 midfielder
ZhangSan18 50 midfielder
ZhangSan19 50 midfielder
ZhangSan20 50 midfielder
ZhangSan21 50 midfielder
ZhangSan22 50 midfielder
ZhangSan23 50 midfielder
0
impossible
/*
* Author: Joshua
* Created Time: 2014年08月29日 星期五 18时52分48秒
* File Name: hdu3720.cpp
*/
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
using namespace std;
#define inf 100000000
char s[35],ps[35];
map<string,int> mp;
const int limit[5]={0,1,4,4,2};
int v[25],g[25][25],id[25],cnt[5],t[12],ans; int gao(char x[])
{
if (x[0]=='g') return 1;
if (x[0]=='d') return 2;
if (x[0]=='m') return 3;
return 4;
} void init()
{
int m,x;
mp.clear();
memset(g,0,sizeof(g));
for (int i=1;i<=23;++i)
{
mp[s]=i;
id[i]=gao(ps);
if (i<23) scanf("%s%d%s",s,&v[i+1],ps);
}
scanf("%d",&m);
for (int i=1;i<=m;++i)
{
scanf("%s%s%d",s,ps,&x);
g[mp[s]][mp[ps]]=x;
g[mp[ps]][mp[s]]=x;
}
} void updata()
{
int temp=0;
for (int i=1;i<=11;++i)
temp+=v[t[i]];
for (int i=1;i<=11;++i)
for (int j=i+1;j<=11;++j)
temp+=g[t[i]][t[j]];
if (temp>ans) ans=temp;
} void dfs(int x,int y)
{
if (y==11)
{
updata();
return;
}
if (x>23) return;
for (int i=0;i<=1;++i)
{
cnt[id[x]]+=i;
if (cnt[id[x]]<=limit[id[x]])
{ if (i) t[y+1]=x;
dfs(x+1,y+i);
}
cnt[id[x]]-=i;
}
} void solve()
{
memset(cnt,0,sizeof(cnt));
ans=-inf;
dfs(1,0);
if (ans==-inf) printf("impossible\n");
else printf("%d\n",ans);
} int main()
{
while (scanf("%s%d%s",s,&v[1],ps)!=EOF)
{
init();
solve();
}
return 0;
}
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