BZOJ1449: [JSOI2009]球队收益

BZOJ1449: [JSOI2009]球队收益

https://lydsy.com/JudgeOnline/problem.php?id=1280

分析:

• 和剪刀石头布那题建图很像，但这题输也有贡献，于是我们令他一开始全输就好了。
• 建图基本都一样就不说了，因为\(D_i\le C_i\)所以费用递增。

代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define N 7500
#define M 2000050
typedef double f2;
namespace EK {
    const int S=N-1,T=N-2;
    int head[N],to[M],nxt[M],flow[M],val[M],cnt=1;
    int dis[N],Q[N],path[N],vis[N];
    inline void add(int u,int v,int f,int c) {
        to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt; flow[cnt]=f; val[cnt]=c;
        to[++cnt]=u; nxt[cnt]=head[v]; head[v]=cnt; flow[cnt]=0; val[cnt]=-c;
    }
    bool spfa() {
        memset(path,0,sizeof(path));
        memset(dis,0x3f,sizeof(dis));
        int l=0,r=0;
        Q[r++]=S; dis[S]=0;
        while(l!=r) {
            int x=Q[l++]; if(l==S) l=0;
            int i; vis[x]=0;
            for(i=head[x];i;i=nxt[i]) if(dis[to[i]]>dis[x]+val[i]&&flow[i]) {
                dis[to[i]]=dis[x]+val[i]; path[to[i]]=i^1;
                if(!vis[to[i]]) {
                    vis[to[i]]=1; Q[r++]=to[i]; if(r==S) r=0;
                }
            }
        }
        return path[T]!=0;
    }
    pair<int,int> ek() {
        int minc=0,maxf=0;
        while(spfa()) {
            int nf=inf;
            int i;
            for(i=T;i!=S;i=to[path[i]]) {
                nf=min(nf,flow[path[i]^1]);
            }maxf+=nf;
            for(i=T;i!=S;i=to[path[i]]) {
                flow[path[i]^1]-=nf;
                flow[path[i]]+=nf;
                minc+=nf*val[path[i]^1];
            }
        }
        return make_pair(minc,maxf);
    }
}
int n,m,wi[N],lo[N],tot[N],C[N],D[N],K;
int Abs(int x) {return x>0?x:-x;}
int pf(int x) {return x*x;}
int calc(int x,int y) {
    return pf(wi[x]+y)*C[x]+pf(lo[x]+tot[x]-y)*D[x];
}
int main() {
    using namespace EK;
    scanf("%d%d",&n,&m);
    int i,j,x,y;
    K=n;
    for(i=1;i<=n;i++) scanf("%d%d%d%d",&wi[i],&lo[i],&C[i],&D[i]);
    for(i=1;i<=m;i++) {
        scanf("%d%d",&x,&y); tot[x]++; tot[y]++;
        K++; add(K,x,1,0); add(K,y,1,0); add(S,K,1,0);
    }
    for(i=1;i<=n;i++) {
        for(j=1;j<=tot[i];j++) {
            add(i,T,1,calc(i,j)-calc(i,j-1));
        }
    }
    int sum=0;
    for(i=1;i<=n;i++) sum+=pf(tot[i]+lo[i])*D[i]+pf(wi[i])*C[i];
    printf("%d\n",ek().first+sum);
}