POJ 2227 The Wedding Juicer——堆的应用——Pku2227

这是刘汝佳神牛《算法艺术与信息学竞赛》上有关堆的一道例题。主要思路如下：

维护一个堆，存储的是水位和高度一样的格子的高度。

1、首先把边缘点全部加到堆中

2、每次取堆中最小的一个，做一次Floodfill，如果被fill到的点比队首元素高度高，那么将这个点加入堆中，否则将该点的水位设置为队首元素的水位。

3、重复2，直到堆中没有元素。

4、最后，统计答案为每个点水位高度与实际高度的差值和。

CODE

  1 const
  2     maxn=90000;
  3     maxp=300;
  4     ddx:array[1..4]of integer=(0,0,1,-1);
  5     ddy:array[1..4]of integer=(1,-1,0,0);
  6 type
  7     rec=record
  8         x,y,data:longint;
  9         end;
 10 var
 11     i,j,m,n,ans                    :longint;
 12     map,high                    :array[1..maxp,1..maxp]of longint;
 13     v                            :array[1..maxp,1..maxp]of boolean;
 14     heap_size                    :longint;
 15     heap                        :array[1..maxn]of rec;
 16     tmp                            :rec;
 17 
 18 procedure down(i:longint);
 19 var j:longint;tmp:rec;
 20 begin
 21     while (i<<1)<=heap_size do
 22         begin
 23             j:=i << 1;
 24             if (heap[j].data>heap[j+1].data)and(j<heap_size)then inc(j);
 25             if heap[j].data<heap[i].data then
 26                 begin
 27                     tmp:=heap[i];heap[i]:=heap[j];heap[j]:=tmp;
 28                     i:=j;
 29                 end else break;
 30         end;
 31 end;
 32 
 33 procedure up(i:longint);
 34 var j:longint;tmp:rec;
 35 begin
 36     while i>1 do
 37         begin
 38             j:=i >> 1;
 39             if heap[j].data>heap[i].data then
 40                 begin
 41                     tmp:=heap[i];heap[i]:=heap[j];heap[j]:=tmp;
 42                     i:=j;
 43                 end else break;
 44         end;
 45 end;
 46 
 47 procedure build_heap;
 48 begin
 49     for i:=(heap_size>>1)downto 1 do down(i);
 50 end;
 51 
 52 procedure floodfill(p:rec);
 53 var x,y,h,t,i                :longint;
 54     seq                        :array[1..maxn]of rec;
 55 begin    
 56     h:=0;t:=1;seq[1]:=p;
 57     while h<t do
 58         begin
 59             inc(h);
 60             x:=seq[h].x;y:=seq[h].y;
 61             for i:=1 to 4 do 
 62                 if (x+ddx[i]>0)and(y+ddy[i]>0)and(x+ddx[i]<=m)and(y+ddy[i]<=n)then 
 63                     begin    
 64                         if (not v[x+ddx[i],y+ddy[i]]) then 
 65                             begin
 66                                 v[x+ddx[i],y+ddy[i]]:=true;
 67                                 if map[x+ddx[i],y+ddy[i]]<=high[x,y] then 
 68                                     begin
 69                                         high[x+ddx[i],y+ddy[i]]:=high[x,y];
 70                                         inc(t);seq[t].x:=x+ddx[i];
 71                                         seq[t].y:=y+ddy[i];
 72                                         seq[t].data:=map[x+ddx[i],y+ddy[i]];
 73                                     end else 
 74                                     begin
 75                                         inc(heap_size);
 76                                         heap[heap_size].x:=x+ddx[i];heap[heap_size].y:=y+ddy[i];
 77                                         heap[heap_size].data:=high[x+ddx[i],y+ddy[i]];
 78                                         up(heap_size);
 79                                     end;
 80                             end;
 81                     end;
 82         end;
 83 end;
 84 
 85 begin
 86     heap_size:=0;
 87     readln(n,m);
 88     for i:=1 to m do
 89         for j:=1 to n do
 90             begin
 91                 read(map[i,j]);
 92                 high[i,j]:=map[i,j];
 93             end;
 94     fillchar(v,sizeof(v),0);
 95     for i:=1 to m do
 96         begin
 97             v[i,1]:=true;inc(heap_size);heap[heap_size].x:=i;heap[heap_size].y:=1;heap[heap_size].data:=map[i,1];
 98             v[i,n]:=true;inc(heap_size);heap[heap_size].x:=i;heap[heap_size].y:=n;heap[heap_size].data:=map[i,n];
 99         end;
100     for j:=2 to n-1 do
101         begin
102             v[1,j]:=true;inc(heap_size);heap[heap_size].x:=1;heap[heap_size].y:=j;heap[heap_size].data:=map[1,j];
103             v[m,j]:=true;inc(heap_size);heap[heap_size].x:=m;heap[heap_size].y:=j;heap[heap_size].data:=map[m,j];
104         end;
105     build_heap;
106     while heap_size>0 do 
107         begin
108             floodfill(heap[1]);
109             tmp:=heap[1];heap[1]:=heap[heap_size];heap[heap_size]:=tmp;
110             dec(heap_size);
111             if heap_size>0 then down(1);
112         end;
113     ans:=0;
114     for i:=1 to m do for j:=1 to n do inc(ans,high[i,j]-map[i,j]);
115     writeln(ans);
116 end.