bzoj 1067 分情况讨论

 

这道题考察人的严谨,各种情况分类讨论。

 

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <map>
 4 #define maxn 50010
 5 #define maxq 17
 6 #define P(i) (1<<(i))
 7 #define yes     0
 8 #define no         1
 9 #define maybe     2
10 using namespace std;
11 
12 int n, q;
13 int year[maxn];
14 int st[maxn][maxq];
15 
16 int ind( int y ) {
17     return (lower_bound( year+1, year+n+1, y ) - year);
18 }
19 void makest() {
20     for( int q=1; q<maxq; q++ ) 
21         for( int i=1; i<=(n-P(q)+1); i++ ) 
22             st[i][q] = max( st[i][q-1], st[i+P(q-1)][q-1] );
23 }
24 int fmax( int lf, int rg ) {
25     int len = rg-lf+1;
26     if( len<=0 ) return -1;
27     for( int q=maxq-1; q>=0; q-- ) 
28         if( len>P(q) ) {
29             return max( st[lf][q], st[rg-P(q)+1][q] );
30         } else if( len==P(q) ) {
31             return st[lf][q];
32         }
33     return -1;
34 }
35 int main() {
36     scanf( "%d", &n );
37     year[n+1] = 0x7FFFFFFF;
38     for( int i=1; i<=n; i++ ) 
39         scanf( "%d%d", year+i, &st[i][0] );
40     makest();
41     sort( year+1, year+1+n );
42     scanf( "%d", &q );
43     while( q-- ) {
44         int a, b, ia, ib;
45         scanf( "%d%d", &a, &b );
46         ia = ind(a), ib = ind(b);
47         if( year[ia]==a && year[ib]==b && b-a==ib-ia && st[ia][0]>=st[ib][0] && fmax(ia+1,ib-1)<st[ib][0] ) {
48             printf( "true\n" );
49             continue;
50         }
51         if( year[ia]==a && year[ib]==b && b-a>ib-ia && st[ia][0]>=st[ib][0] && fmax(ia+(year[ia]==a),ib-1)<st[ib][0] ) {
52             printf( "maybe\n" );
53             continue;
54         }
55         if( ((year[ia]==a)^(year[ib]==b)) && fmax(ia+(year[ia]==a),ib-1)< (year[ia]==a ? st[ia][0] : st[ib][0])  ) {
56             printf( "maybe\n" );
57             continue;
58         }
59         if( year[ia]!=a && year[ib]!=b ) {
60             printf( "maybe\n" );
61             continue;
62         }
63         printf( "false\n" );
64     }
65 }
View Code

 

posted @ 2015-02-06 23:03  idy002  阅读(113)  评论(0编辑  收藏  举报