Pku3171 cleaning shifts

There is a warehouse. [M, E] needs to be cleaned all the time. Now there are native cows to do the job. Each cow has a working period [start, final], and the cost is cost. The question is, which cows can be hired at least? Multiple cows can work at the same time.

First, sort the intervals in ascending order of final and then perform DP. Considering the large range and O (L ^ 2) Timeout, we introduce the line segment tree for optimization and reduce the decision complexity to logn.

And output-1, WA is dead...

This is my PKU 100th question. I'm so excited to celebrate ..

View code

  1 program pku3171(input,output);  2 type  3     node=record  4         left,right,x,y:longint;  5         minn:longint;  6     end;  7 var  8     tree:array[0..180000] of node;  9     f    :array[0..90000] of longint; 10     start,final,cost:array[0..11000] of longint; 11     n,m,e:longint; 12     tot:longint; 13     maxl:longint; 14 function min(aa,bb:longint):longint; 15 begin 16     if aa<bb then 17         exit(aa); 18     exit(bb); 19 end;{ min } 20 procedure build(xx,yy:longint); 21 var 22     now,mid:longint; 23 begin 24     inc(tot); 25     now:=tot; 26     tree[now].x:=xx; 27     tree[now].y:=yy; 28     if xx=yy then 29     begin 30         tree[now].minn:=19950714; 31         exit; 32     end; 33     mid:=(xx+yy)>>1; 34     tree[now].left:=tot+1; 35     build(xx,mid); 36     tree[now].right:=tot+1; 37     build(mid+1,yy); 38     tree[now].minn:=19950714; 39 end;{ build } 40 procedure change(now,poss,worth:longint); 41 var 42     mid:longint; 43 begin 44     if tree[now].x=tree[now].y then 45     begin 46         tree[now].minn:=worth; 47         exit; 48     end; 49     mid:=(tree[now].x+tree[now].y)>>1; 50     if poss<=mid then 51         change(tree[now].left,poss,worth) 52     else 53         change(tree[now].right,poss,worth); 54     tree[now].minn:=min(tree[tree[now].left].minn,tree[tree[now].right].minn); 55 end;{ change } 56 function find(now,xx,yy:longint):longint; 57 var 58     mid:longint; 59 begin 60     if (tree[now].x=xx)and(tree[now].y=yy) then 61         exit(tree[now].minn); 62     mid:=(tree[now].x+tree[now].y)>>1; 63     if yy<=mid then 64         exit(find(tree[now].left,xx,yy)) 65     else 66         if xx>mid then 67             exit(find(tree[now].right,xx,yy)) 68         else 69             exit(min(find(tree[now].left,xx,mid),find(tree[now].right,mid+1,yy))); 70 end;{ find } 71 procedure swap(var aa,bb:longint); 72 var 73     tt:longint; 74 begin 75     tt:=aa; 76     aa:=bb; 77     bb:=tt; 78 end;{ swap } 79 procedure sort(p,q:longint); 80 var 81     i,j,mid:longint; 82 begin 83     i:=p; 84     j:=q; 85     mid:=final[(i+j)>>1]; 86     repeat 87         while final[i]<mid do 88             inc(i); 89         while final[j]>mid do 90             dec(j); 91         if i<=j then 92         begin 93             swap(start[i],start[j]); 94             swap(final[i],final[j]); 95             swap(cost[i],cost[j]); 96             inc(i); 97             dec(j); 98         end; 99     until i>j;100     if i<q then sort(i,q);101     if j>p then sort(p,j);102 end;{ sort }103 procedure init;104 var105     i:longint;106 begin107     readln(n,m,e);108     tot:=0;109     maxl:=0;110     for i:=1 to n do111     begin112         readln(start[i],final[i],cost[i]);113         if final[i]>maxl then114             maxl:=final[i];115     end;116     build(0,maxl);117     sort(1,n);118 end;{ init }119 function max(aa,bb:longint):longint;120 begin121     if aa>bb then122         exit(aa);123     exit(bb);124 end;{ max}125 procedure main;126 var127     i,j:longint;128 begin129     fillchar(f,sizeof(f),63);130     change(1,m,0);131     f[m]:=0;132     for i:=1 to n do133     begin134         j:=find(1,max(start[i]-1,0),final[i]);135         if j+cost[i]<f[final[i]] then136         begin137             f[final[i]]:=j+cost[i];138             change(1,final[i],f[final[i]]);139         end;140     end;141 end;{ main }142 procedure print;143 begin144     if f[e]>=19950714 then145         writeln(-1)146     else147         writeln(f[e]);148 end;{ print }149 begin150     init;151     main;152     print;153 end.