HDU 3086 马拉车模板

模板，但是对这个算法还是不太清楚,真实不明觉厉....

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string.h>
 4 #pragma warning ( disable : 4996 )
 5 using namespace std;
 6 
 7 inline int Max(int a,int b) { return a>b?a:b; }
 8 inline int Min(int a,int b) { return a>b?b:a; }
 9 const int inf = 0x3f3f3f3f;
10 const int maxn = 1e5+1e4+5;
11 
12 char str[maxn];
13 char nstr[maxn<<1];
14 int maxlen[maxn<<1];
15 int len, plen, ans;
16 
17 void init()
18 {
19     memset( nstr, 0, sizeof(nstr) );
20     memset( maxlen, 0, sizeof(maxlen) );
21     len = strlen(str);    
22     nstr[0] = '$'; nstr[1] = '#';
23 
24     int j = 2;
25     for ( int i = 0; i < len; i++ )
26     {
27         nstr[j++] = str[i];
28         nstr[j++] = '#';
29     }
30     nstr[j] = '\0';
31     plen = j;
32 }
33 
34 void solve()
35 {
36     ans = -1;
37     int id, mx = 0;
38 
39     for ( int i = 1; i < plen; i++ )
40     {
41         if( i < mx )
42             maxlen[i] = Min( maxlen[2*id-i], mx-i );
43         else
44             maxlen[i] = 1;
45 
46         while( nstr[i-maxlen[i]] == nstr[i+maxlen[i]] )
47             maxlen[i]++;
48 
49         if ( mx < i + maxlen[i] )
50         {
51             id = i;
52             mx = i + maxlen[i];
53         }
54         ans = Max(ans, maxlen[i]-1);
55     }
56 }
57 
58 int main()
59 {
60     while ( ~scanf("%s", str) )
61     {
62         init();
63         solve();
64         printf( "%d\n", ans );
65     }
66     return 0;
67 }
View Code

 又做了一道几乎模板的题（吉哥系列故事——完美队形II），终于对马拉车有点理解了，这算法实在太巧妙了！

和模板几乎一样，只不过增多了个左半边要升序排列罢了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string.h>
 4 #pragma warning ( disable : 4996 )
 5 using namespace std;
 6 
 7 inline int Max(int a,int b) { return a>b?a:b; }
 8 inline int Min(int a,int b) { return a>b?b:a; }
 9 const int inf = 0x3f3f3f3f;
10 const int maxn = 1e5+5;
11 
12 int maxl[maxn<<1];
13 int str[maxn], nstr[maxn<<1];
14 int N, ans, plen;
15 
16 void init()
17 {
18     memset( maxl, 0, sizeof(maxl) );
19     memset( nstr, 0, sizeof(nstr) );
20     nstr[0] = 0; nstr[1] = -1;
21 
22     int j = 2;
23     for( int i = 0; i < N; i++ )
24         { nstr[j++] = str[i]; nstr[j++] = -1; }
25     nstr[j] = -2;
26     plen = j;
27 }
28 
29 void solve()
30 {
31     ans = -1;
32     int id, mx = 0;
33     for ( int i = 1; i < plen; i++ )
34     {
35         if(i < mx) 
36             maxl[i] = Min(maxl[2*id-i], mx-i);
37         else 
38             maxl[i] = 1;
39         
40         while( nstr[i-maxl[i]]==nstr[i+maxl[i]] && nstr[i-maxl[i]]<=nstr[i-maxl[i]+2] )
41             maxl[i]++;
42         if(mx < i + maxl[i])
43             { id = i; mx = i + maxl[i]; }
44 
45         ans = Max(ans, maxl[i]-1);
46     }
47 }
48 
49 int main()
50 {
51     int all; cin >> all;
52     while (all--)
53     {
54         cin >> N;
55         for( int i = 0; i < N; i++ )
56             scanf( "%d", &str[i] );
57         init();
58         solve();
59 
60         printf( "%d\n", ans );
61     }
62     return 0;
63 }
View Code

 

posted @ 2018-03-21 21:30  LBNOQYX  阅读(136)  评论(0)  编辑  收藏  举报