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 }
