bzoj 1047 单调队列

做4次单调队列优化DP。

 1 /**************************************************************
 2     Problem: 1047
 3     User: idy002
 4     Language: C++
 5     Result: Accepted
 6     Time:2088 ms
 7     Memory:16756 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #define N 1010
12
13 struct Pair {
14     int i, j, v;
15     Pair(){}
16     Pair( int i, int j, int v ):i(i),j(j),v(v){}
17 };
18
19 int n, m, r, ans;
20 int v[N][N], dp[N][N], dmin[N][N], dmax[N][N];
21 Pair qu[N]; int bg, ed;
22
23 int dodp() {
24     //  min
25     for( int i=1; i<=n; i++ ) {
26         bg=1, ed=0;
27         for( int j=1; j<=m; j++ ) {
28             while( bg<=ed && qu[bg].j<j-r+1 ) bg++;
29             while( bg<=ed && qu[ed].v>v[i][j] ) ed--;
30             qu[++ed] = Pair(i,j,v[i][j]);
31             dp[i][j] = qu[bg].v;
32         }
33     }
34     for( int j=1; j<=m; j++ ) {
35         bg=1, ed=0;
36         for( int i=1; i<=n; i++ ) {
37             while( bg<=ed && qu[bg].i<i-r+1 ) bg++;
38             while( bg<=ed && qu[ed].v>dp[i][j] ) ed--;
39             qu[++ed] = Pair(i,j,dp[i][j]);
40             dmin[i][j] = qu[bg].v;
41         }
42     }
43     //  max
44     for( int i=1; i<=n; i++ ) {
45         bg=1, ed=0;
46         for( int j=1; j<=m; j++ ) {
47             while( bg<=ed && qu[bg].j<j-r+1 ) bg++;
48             while( bg<=ed && qu[ed].v<v[i][j] ) ed--;
49             qu[++ed] = Pair(i,j,v[i][j]);
50             dp[i][j] = qu[bg].v;
51         }
52     }
53     for( int j=1; j<=m; j++ ) {
54         bg=1, ed=0;
55         for( int i=1; i<=n; i++ ) {
56             while( bg<=ed && qu[bg].i<i-r+1 ) bg++;
57             while( bg<=ed && qu[ed].v<dp[i][j] ) ed--;
58             qu[++ed] = Pair(i,j,dp[i][j]);
59             dmax[i][j] = qu[bg].v;
60         }
61     }
62     int ans = 2000000001;
63     for( int i=r; i<=n; i++ )
64         for( int j=r; j<=m; j++ ) {
65             int tans = dmax[i][j]-dmin[i][j];
66             if( ans>tans ) ans=tans;
67         }
68     /*  debug
69     fprintf( stderr, "n=%d m=%d r=%d\n", n, m, r );
70     fprintf( stderr, "v:\n" );
71     for( int i=1; i<=n; i++ ) {
72         for( int j=1; j<=m; j++ )
73             fprintf( stderr, "%2d ", v[i][j] );
74         fprintf( stderr, "\n" );
75     }
76     fprintf( stderr, "dmin:\n" );
77     for( int i=1; i<=n; i++ ) {
78         for( int j=1; j<=m; j++ )
79             fprintf( stderr, "%2d ", dmin[i][j] );
80         fprintf( stderr, "\n" );
81     }
82     fprintf( stderr, "dmax:\n" );
83     for( int i=1; i<=n; i++ ) {
84         for( int j=1; j<=m; j++ )
85             fprintf( stderr, "%2d ", dmax[i][j] );
86         fprintf( stderr, "\n" );
87     }
88     */
89     return ans;
90 }
91 int main() {
92     scanf( "%d%d%d", &n, &m, &r );
93     for( int i=1; i<=n; i++ )
94         for( int j=1; j<=m; j++ )
95             scanf( "%d", &v[i][j] );
96     printf( "%d\n", dodp() );
97 }