牛客多校第五场J

题意：有n颗钻石，位于三维坐标，每秒下沉vi，人每秒可捞回一棵钻石，捞回一棵钻石的代价为到原点距离的平方，问最小代价

解：肯定是0-n-1时间内捞回n颗钻石，所以是一张带权的二分图，二分图最小权匹配板子题

下附代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long
using namespace std;
const int N = 305;
const ll inf = 1e18;
struct node
{
    ll x, y, z, v;
}p[N];
ll n,m,val[N][N],matched[N];
ll slack[N],pre[N],ex[N],ey[N];//ex,ey顶标
bool visx[N],visy[N];
void match(ll u)
{
    ll x,y=0,yy=0,delta;
    memset(pre,0,sizeof(pre));
    for(ll i=1;i<=n;i++)slack[i]=inf;
    matched[y]=u;
    while(1)
    {
        x=matched[y];delta=inf;visy[y]=1;
        for(ll i=1;i<=n;i++)
        {
            if(visy[i])continue;
            if(slack[i]>ex[x]+ey[i]-val[x][i])
            {
                slack[i]=ex[x]+ey[i]-val[x][i];
                pre[i]=y;
            }
            if(slack[i]<delta){delta=slack[i];yy=i;}
        }
        for(ll i=0;i<=n;i++)
        {
            if(visy[i])ex[matched[i]]-=delta,ey[i]+=delta;
            else slack[i]-=delta;
        }
        y=yy;
        if(matched[y]==-1)break;
    }
    while(y){matched[y]=matched[pre[y]];y=pre[y];}
}
ll KM()
{
    memset(matched,-1,sizeof(matched));
    memset(ex,0,sizeof(ex));
    memset(ey,0,sizeof(ey));
    for(ll i=1;i<=n;i++)
    {
        memset(visy,0,sizeof(visy));
        match(i);
    }
    ll res=0;
    for(ll i=1;i<=n;i++)
        if(matched[i]!=-1)res+=val[matched[i]][i];
    return res;
}

int main()
{
    scanf("%lld",&n);
        for (ll i = 1; i <=n; ++i){
             ll x, y, z, v;
            scanf("%lld%lld%lld%lld", &x, &y, &z, &v);
            for (ll j = 1; j <=n; ++j)
               val[i][j]=-(x*x+y*y+(z+(j-1)*v)*(z+(j-1)*v));
        }
        printf("%lld", -KM());
    return 0;
}