topcoder FoxAndCity

来源：互联网发布：知与行杂志版面费编辑：程序博客网时间：2024/06/06 01:24

可以看成给每个点随便选权值，有相邻边的点权值相差不超过1。因为原图是联通的，每种满足要求的方案都一定是合法的。这样建图跑最小割就行了，参见【HNOI2013】bzoj3144 切糕。

#include<algorithm>#include<vector>#include<string>using namespace std;const int maxn=200010,s=20005,t=20006,oo=0x3f3f3f3f;class FoxAndCity{    int fir[maxn],ne[maxn],to[maxn],w[maxn],    que[maxn],dep[maxn],cur[maxn],id[45][45],    n,num,tot;private:    void add(int u,int v,int x)    {        num++;        ne[num<<1]=fir[u];        fir[u]=num<<1;        to[num<<1]=v;        w[num<<1]=x;        ne[num<<1|1]=fir[v];        fir[v]=num<<1|1;        to[num<<1|1]=u;        w[num<<1|1]=0;    }    int bfs()    {        int hd=1,tl=1,u,v;        for (int i=1;i<=tot;i++) dep[i]=0;        dep[t]=0;        que[dep[s]=1]=s;        while (hd<=tl)        {            u=que[hd++];            cur[u]=fir[u];            for (int i=fir[u];i;i=ne[i])                if (w[i]&&!dep[v=to[i]])                {                    dep[v]=dep[u]+1;                    que[++tl]=v;                }        }        return dep[t];    }    int dfs(int u,int lim)    {        if (u==t) return lim;        int ret=0,v,x;        for (int &i=cur[u];i&&ret<lim;i=ne[i])            if (w[i]&&dep[v=to[i]]==dep[u]+1)            {                x=dfs(v,min(w[i],lim-ret));                ret+=x;                w[i]-=x;                w[i^1]+=x;            }        return ret;    }public:    int minimalCost(vector<string> linked,vector<int> want)    {        int ans=0;        n=want.size();        tot=num=0;        for (int i=0;i<n;i++)            for (int j=0;j<=n;j++)                id[i][j]=++tot;        for (int i=1;i<=tot;i++) fir[i]=0;        fir[s]=fir[t]=0;        add(s,id[0][0],oo);        add(id[0][n],t,oo);        for (int i=1;i<n;i++) add(id[0][i],id[0][i+1],oo);        for (int i=1;i<n;i++)        {            add(s,id[i][0],oo);            add(id[i][0],id[i][1],oo);            add(id[i][n],t,oo);            for (int j=want[i]-1;j>=1;j--)                add(id[i][j],id[i][j+1],(want[i]-j)*(want[i]-j));            for (int j=want[i]+1;j<n;j++)                add(id[i][j],id[i][j+1],(j-want[i])*(j-want[i]));        }        for (int i=0;i<n;i++)            for (int j=i+1;j<n;j++)                if (linked[i][j]=='Y')                    for (int k=0;k<n;k++)                    {                        add(id[i][k+1],id[j][k],oo);                        add(id[j][k+1],id[i][k],oo);                    }        while (bfs()) ans+=dfs(s,oo);        return ans;    }};

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