• bzoj2595[Wc2008]游览计划


    http://www.lydsy.com/JudgeOnline/problem.php?id=2595

    斯坦纳树。

    斯坦纳树是在求一个图中的包含某些特定点的最小生成树,注意其他点不一定要包含。

    这道题可以算是斯坦纳树的入门题了。

    题解请戳

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<fstream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<map>
    #include<utility>
    #include<set>
    #include<bitset>
    #include<vector>
    #include<functional>
    #include<deque>
    #include<cctype>
    #include<climits>
    #include<complex>
    //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj
     
    using namespace std;
    
    typedef long long LL;
    typedef double DB;
    typedef pair<int,int> PII;
    typedef complex<DB> CP;
    
    #define mmst(a,v) memset(a,v,sizeof(a))
    #define mmcy(a,b) memcpy(a,b,sizeof(a))
    #define fill(a,l,r,v) fill(a+l,a+r+1,v)
    #define re(i,a,b)  for(i=(a);i<=(b);i++)
    #define red(i,a,b) for(i=(a);i>=(b);i--)
    #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
    #define fi first
    #define se second
    #define m_p(a,b) make_pair(a,b)
    #define p_b(a) push_back(a)
    #define SF scanf
    #define PF printf
    #define two(k) (1<<(k))
    
    template<class T>inline T sqr(T x){return x*x;}
    template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
    template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}
    
    const DB EPS=1e-9;
    inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
    const DB Pi=acos(-1.0);
    
    inline int gint()
      {
            int res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    inline LL gll()
      {
          LL res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    
    const int maxN=10;
    const int maxK=10;
    const int INF=0xf0f0f0f;
    const int dx[]={1,-1,0,0};
    const int dy[]={0,0,1,-1};
    
    int N,M,K;
    int a[maxN+10][maxN+10];
    int id[maxN+10][maxN+10];
    
    int F[maxN+10][maxN+10][two(maxK)+100];
    struct Tpre
      {
          int x,y,s;
          inline Tpre(int _x=0,int _y=0,int _s=0){x=_x;y=_y;s=_s;}
      }pre[maxN+10][maxN+10][two(maxK)+100];
    
    inline int update(int x,int y,int state,int px,int py,int pstate,int v)
      {
          if(v<F[x][y][state]) return F[x][y][state]=v,pre[x][y][state]=Tpre(px,py,pstate),1;
          return 0;
      }
    
    queue<PII>Q;
    int vis[maxN+10][maxN+10];
    
    inline void SPFA(int state)
      {
          while(!Q.empty())
            {
                int i,x=Q.front().fi,y=Q.front().se;Q.pop();
                vis[x][y]=0;
                re(i,0,3)
                  {
                      int _x=x+dx[i],_y=y+dy[i];
                      if(_x<1 || _x>N || _y<1 || _y>M)continue;
                      if(update(_x,_y,state,x,y,state,F[x][y][state]+a[_x][_y]) && !vis[_x][_y])vis[_x][_y]=1,Q.push(PII(_x,_y));
                  }
            }
      }
    
    inline void DFS(int i,int j,int state)
      {
          if(i==0 && j==0 && state==0)return;
          vis[i][j]=1;
          int x=pre[i][j][state].x,y=pre[i][j][state].y,s=pre[i][j][state].s;
          DFS(x,y,s);
          if(i==x && j==y)DFS(x,y,state-s);
      }
    
    inline void output()
      {
          int i,j;
            re(i,1,N)
              {
                  re(j,1,M)
                        if(a[i][j]==0)
                          putchar('x');
                        else
                        if(vis[i][j])
                          putchar('o');
                        else
                          putchar('_');
                    putchar('
    ');
                }
        }
    
    int main()
      {
          freopen("bzoj2595.in","r",stdin);
          freopen("bzoj2595.out","w",stdout);
          int i,j;
          N=gint();M=gint();
          mmst(F,0xf);
          re(i,1,N)re(j,1,M)
              {
                  a[i][j]=gint();
                    if(a[i][j]==0)id[i][j]=++K,F[i][j][two(K-1)]=0;
                }
            int state,maxstate=two(K)-1;
            re(state,1,maxstate)
              {
                  re(i,1,N)re(j,1,M)
                    {
                      for(int s=state&(state-1);s;s=(s-1)&state)
                        update(i,j,state,i,j,s,F[i][j][s]+F[i][j][state-s]-a[i][j]);
                            if(F[i][j][state]!=INF)Q.push(PII(i,j)),vis[i][j]=1;
                    }
                  SPFA(state);
              }
            re(i,1,N)re(j,1,M)if(a[i][j]==0)
              {
                  cout<<F[i][j][maxstate]<<endl;
                  DFS(i,j,maxstate);
                  output();
                  return 0;
                }
            return 0;
        }
    View Code
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  • 原文地址:https://www.cnblogs.com/maijing/p/4761458.html
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