bzoj1449&&bzoj2895

题解：

S连每场比赛流量1费用0 
每场比赛连参赛队流量1费用0 

我们发现调整一次 由win，lose变为 win+1，lose-1的费用为 
(C*(win+1)^2+D*(lose-1)^2) - (C*win^2+D*lose^2)=C*(2*win+1)-D*(2*lose-1) 
暴力连边就可以了 
最后的答案=最想费用流+最初假设所有队伍都输的收益

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=100005;
int S,W,lose[N],sum[N],win[N],ans,C[N],D[N];
int fi[N],n,t,cas,m,x,y,z,f[N],ne[N],num,zz[N],fl[N],gp[N],dist[N],pre[N],sl[N];
void jb(int x,int y,int z,int s)
{
    ne[num]=fi[x];
    fi[x]=num;
    zz[num]=y;
    sl[num]=z;
    fl[num++]=s;
    ne[num]=fi[y];
    fi[y]=num;
    zz[num]=x;
    sl[num]=0;
    fl[num++]=-s;
}
int spfa()
{
    memset(dist,0x3f,sizeof dist);
    memset(pre,-1,sizeof pre);
    memset(gp,0,sizeof gp);
    memset(f,0,sizeof f);
    queue<int > Q;
    Q.push(S);
    dist[S]=0;
    while (!Q.empty())
     {
         int now=Q.front();
         Q.pop();
         f[now]=0;
         for (int i=fi[now];i!=-1;i=ne[i])
          if (sl[i]>0)
           {
              int t=zz[i];
              if (dist[t]>dist[now]+fl[i])
               {
                   dist[t]=dist[now]+fl[i];
                   pre[t]=now;
                   gp[t]=i;
                   if (!f[t])
                    {
                        f[t]=1;
                        Q.push(t);
                    }
               }
           }
     }
    if (pre[W]==-1)return 1;
    return 0; 
}
void Max_flow()
{
    int cost=0,flow=0;
    while (!spfa())
     {
         int f=1e9;
         for (int i=W;i!=S;i=pre[i])
          f=min(f,sl[gp[i]]);
         cost+=f;
        flow+=dist[W]*f;
        for (int i=W;i!=S;i=pre[i])
         {
             sl[gp[i]]-=f;
             sl[gp[i]^1]+=f;
         } 
     }
    printf("%d\n",flow+ans);
}
int main()
{
    memset(fi,-1,sizeof fi);
    scanf("%d%d",&n,&m); 
    S=n+m+1,W=S+1;
    for (int i=1;i<=n;i++)scanf("%d%d%d%d",&win[i],&lose[i],&C[i],&D[i]);
    for (int i=1;i<=m;i++)
     {
        int x,y;
        scanf("%d%d",&x,&y);
        jb(S,n+i,1,0);
        jb(n+i,x,1,0);
        jb(n+i,y,1,0);
        sum[x]++,sum[y]++;
     }
    for (int i=1;i<=n;i++)lose[i]+=sum[i];
    for (int i=1;i<=n;i++)ans+=C[i]*win[i]*win[i]+D[i]*lose[i]*lose[i];
    for (int i=1;i<=n;i++)
     for (int j=1;j<=sum[i];j++)
      {
        jb(i,W,1,C[i]*(2*win[i]+1)-D[i]*(2*lose[i]-1));
        win[i]++;lose[i]--;
      }
    Max_flow();
}