2020hdu多校round2第五题New Equipments

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=6767

题解

一个工人对应一个二次函数，尽量往函数对称轴配对。而又n个工人，所以每个工人一定能连到对称轴周围n个最近的点之一。所以从n个工人节点向他的对称轴周围n个点连边，边流量1，费用$a_i{j}^2+b_{i}j+c_i$，完了超级源流量限制为k，枚举k建立n次图，跑n次费用流。

AC代码

尽量不用宏定义字符串替换，括号老是丢。找错找一天。。哭了。

#include
#define Pair pair
#define fi first
#define se second
//#define AddEdge(x,y,f,z) add_edge(x,y,f,z);add_edge(y,x,0,-z);
// #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++)
// char buf[1<<20],*p1=buf,*p2=buf;
using namespace std;
const int NN=5010;
long long INF=1000000000000000000ll;
const int MM=50010;
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int N,M,S,T;
struct node
{
    int u,v;
    long long f,w;
}edge[MM];
int num=0;
vector<int>con[NN];
inline void add_edge(int x,int y,long long f,long long z)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].f=f;
    edge[num].w=z;
    con[x].push_back(num);
    num++;
}
inline void AddEdge(int x,int y,long long f,long long z) {add_edge(x,y,f,z);add_edge(y,x,0,-z);}
long long h[NN],dis[NN];
int PrePoint[NN],PreEdge[NN];
Pair Dij()
{
    long long ansflow=0,anscost=0;
    for(int i=1;i<=N;i++)h[i]=0;
    while(1)
    {
        priority_queue<Pair>q;
        for(int i=1;i<=N;i++)dis[i]=INF;
        dis[S]=0;
        q.push(make_pair(0,S));
        while(q.size()!=0)
        {
            Pair p=q.top();q.pop();
            if(-p.fi!=dis[p.se]) continue;
            if(p.se==T) break;
            int up=con[p.se].size();
            for(int ii=0;ii<up;ii++)
            {
                int i=con[p.se][ii];
                long long nowcost=edge[i].w+h[p.se]-h[edge[i].v];
                if(edge[i].f>0&&dis[edge[i].v]>dis[p.se]+nowcost)
                {
                    dis[edge[i].v]=dis[p.se]+nowcost;
                    q.push(make_pair(-dis[edge[i].v],edge[i].v));
                    PrePoint[edge[i].v]=p.se;
                    PreEdge[edge[i].v]=i;
                }
            }
        }
        if(dis[T]>=INF) break;
        for(int i=0;i<=N;i++) h[i]+=dis[i];
        long long nowflow=INF;
        for(int now=T;now!=S;now=PrePoint[now])
            nowflow=min(nowflow,edge[PreEdge[now]].f);
        for(int now=T;now!=S;now=PrePoint[now])
            edge[PreEdge[now]].f-=nowflow,
            edge[PreEdge[now]^1].f+=nowflow;
        ansflow+=nowflow;
        anscost+=nowflow*h[T];
    }
    return make_pair(ansflow,anscost);
}
map<int,int>ma;
long long a[NN],b[NN],c[NN];
vector<int>from[NN];
int rec[NN];
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++){
			scanf("%lld%lld%lld",a+i,b+i,c+i);
		}
		int cnt=0;
		int gs=n/2+n%2;
		for(int i=1;i<=n;i++){
			if(b[i]>0){
				for(int j=1;j<=n;j++){
					if(ma[j]==0)ma[j]=++cnt;
					from[ma[j]].push_back(i);
					rec[ma[j]]=j;
				}
			}
			else{
				int mid=-b[i]/(a[i]<<1);
				if(mid<gs){
					for(int j=1;j<=n;j++){
						if(ma[j]==0)ma[j]=++cnt;
						from[ma[j]].push_back(i);
						rec[ma[j]]=j;
					}
				}
				else{
					if(mid+gs>m){
						for(int j=m-n+1;j<=m;j++){
							if(ma[j]==0)ma[j]=++cnt;
							from[ma[j]].push_back(i);
							rec[ma[j]]=j;
						}
					}
					else{
						for(int j=mid-gs+1;j<=mid+gs;j++){
							if(ma[j]==0)ma[j]=++cnt;
							from[ma[j]].push_back(i);
							rec[ma[j]]=j;
						}
					}
				}
			}
		}
		N=n+3+cnt;
		for(int k=1;k<=n;k++){
			for(int i=1;i<=N;i++){
				con[i].clear();
			}
			num=0;
			for(int i=1;i<=cnt;i++){
				int up=from[i].size();
				int jj=rec[i];
				for(int j=0;j<up;j++){
					int fr=from[i][j];
					AddEdge(fr+2,i+n+2,1,a[fr]*jj*jj+b[fr]*jj+c[fr]);
				}
			}
			for(int i=1;i<=n;i++){
				AddEdge(2,i+2,1,0);
			}
			for(int i=n+2+1;i<=n+2+cnt;i++){
				AddEdge(i,N,1,0);
			}
			AddEdge(1,2,k,0);
			// printf("%d\n",num);
			// for(int i=0;i
			// 	printf("%d %d %lld %lld\n",edge[i].u,edge[i].v,edge[i].f,edge[i].w);
			// }
			S=1;T=N;
			Pair ans=Dij();
			if(k==n)printf("%lld",ans.se);
			else printf("%lld ",ans.se);
		}printf("\n");
		for(int i=1;i<=cnt;i++)from[i].clear();
		ma.clear();
	}
	return 0;
}