S向每个人连一条容量为1 费用为0 的边 , 每个人向职位连容量为1 费用为-point的边,每个职位根据模式向T连费用为0的边.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i = 0; i < N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
int flow = 0;
cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
{
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
int n;
int a[505][11];
string s;
map<int ,string> mp;
int b1[][9]=
{
{0,1,2,1,1,2,1,1,2},
{0,1,2,1,1,3,1,1,1},
{0,1,2,1,1,3,0,0,3},
{0,1,3,0,0,3,1,1,2},
{0,1,3,0,0,4,1,1,1},
{0,1,3,1,1,3,0,0,2},
{0,1,4,1,1,3,0,0,1}
};
struct node
{
string nm;
int po;
bool operator < (const node &x) const
{
return po>x.po;
}
};
vector<node> ans;
int main()
{
// freopen("data.txt","r",stdin);
ios_base::sync_with_stdio(false);
mp[1]="A";
mp[2]="B";
mp[3]="C";
mp[4]="D";
mp[5]="E";
mp[6]="F";
mp[7]="China";
while(cin >> n && n)
{
for(int i=1;i<=n;i++)
{
cin >>s;
for(int j=1;j<=8;j++)
{
cin >> a[i][j];
}
}
ans.clear();
for(int k=1;k<=7;k++)
{
int s = 0;
int t = n+10;
init(n+15);
for(int i=1;i<=n;i++)
{
addedge(s,i,1,0);
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=8;j++)
{
addedge(i,j+n,1,-a[i][j]);
}
}
for(int i=1;i<=8;i++)
{
addedge(i+n,t,b1[k-1][i],0);
}
int cos = 0;
int flow = minCostMaxflow(s,t,cos);
// cout << cos<<endl;
if(flow==11)
{
node nt ;
nt.nm = mp[k];
nt.po=-cos;
ans.push_back(nt);
}
}
sort(ans.begin(),ans.end());
cout <<"Formation "<<ans[0].nm<<" has the highest points "<<ans[0].po<<"."<<endl;
for(int i=1;i<ans.size();i++)
{
if(ans[i].po==ans[i-1].po)
cout <<"Formation "<<ans[i].nm<<" has the highest points "<<ans[i].po<<"."<<endl;
else
break;
}
cout << endl;
}
return 0;
}