不难发现必然是两个人之间话产生矛盾或自身话有问题
很显然，当ai>aj时，若ai<n-bj时i,j两人话矛盾
ai<ai<n-bj，这东西有没有数轴上的线段的既视感？
我们只要求出做多不相交的线段就可得到最少谎话数
1 var a,b,f:array[0..100010] of longint; 2 i,j,k,n:longint; 3 4 function min(a,b:longint):longint; 5 begin 6 if a>b then exit(b) else exit(a); 7 end; 8 9 function max(a,b:longint):longint; 10 begin 11 if a>b then exit(a) else exit(b); 12 end; 13 14 procedure swap(var a,b:longint); 15 var c:longint; 16 begin 17 c:=a; 18 a:=b; 19 b:=c; 20 end; 21 22 procedure sort(l,r:longint); 23 var i,j,x,y:longint; 24 begin 25 i:=l; 26 j:=r; 27 x:=b[(l+r) shr 1]; 28 y:=a[(l+r) shr 1]; 29 repeat 30 while (b[i]<x) or (b[i]=x) and (a[i]<y) do inc(i); 31 while (x<b[j]) or (b[j]=x) and (y<a[j]) do dec(j); 32 if not(i>j) then 33 begin 34 swap(a[i],a[j]); 35 swap(b[i],b[j]); 36 inc(i); 37 dec(j); 38 end; 39 until i>j; 40 if l<j then sort(l,j); 41 if i<r then sort(i,r); 42 end; 43 44 begin 45 readln(n); 46 for i:=1 to n do 47 begin 48 readln(a[i],b[i]); 49 b[i]:=n-b[i]; 50 end; 51 sort(1,n); 52 j:=1; 53 for i:=1 to n do 54 begin 55 f[i]:=f[i-1]; 56 while b[j]=i do 57 begin 58 k:=j+1; 59 while (a[k]=a[j]) and (b[k]=b[j]) do inc(k); 60 f[i]:=max(f[i],f[a[j]]+min(k-j,b[j]-a[j])); //注意这个地方相同分数的数目 61 j:=k; 62 end; 63 end; 64 writeln(n-f[n]); 65 end.
