UVa 311-packets

Topic: There are 1x1,2x2,3x3,4x4,5x5,6x6 of square blocks, ask at least a few 6x6 boxes can be put down.

Analysis: Greedy heart. 1x1 is used to fill the remaining space, any space can be put down;

First of all, each 6x6,5x5,4x4 to a single box, so the rest of the space to fill the 2x2,1x1 as the best;

Then, the rest of the 3x3,2x2, first filled with the placement (4 3x3,9 2x2 Group assigned to the best);

Finally, there are less than 4 3x3, 2x2 less than 9, up to two boxes, 1x1 last fill.

(proving that the second and third steps are optimal, with a total of 36 combinations, of which 2 boxes are:

<1,6>,<1,7>,<1,8>,<1,6>,

<2,4>,<2,5>,<2,6>,<2,7>,<2,8>

<3,2>,<3,3>,<3,4>,<3,5>,<3,6>,<3,7>,<3,8>

For each situation, add <4,9> redistribute, you can't get smaller boxes, so they are best allocated.

SAY: See is not see is greedy heart, use DP solution 3x3 and 2x2 allocation way ╮(╯▽╰)╭.

#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include < Cstdio> #include <cmath>using namespace Std;int block[7];int maps[5][10] = { 0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,1,1,1,1,2,2,2,2,2,2,1,1,2,2,2,2,2,2,2,2,1,2,2,2,2,2,2,2,2,2};int dp[505 ][505];int Main () {while (~SCANF ("%d", &block[1]) {int sum = block[1];for (int i = 2; I <= 6; + + i) {scanf ("%d", &amp ; block[i]); sum + = Block[i];} if (!sum) break;//6x6int count = Block[6];//5x5count + = block[5];block[1] = min (block[1], 11*block[5]);//4x4count + = Bloc K[4];if (5*block[4] > Block[2]) {block[1] = min (block[1], 20*block[4]-4*block[2]); block[2] = 0;} else {block[2]-= 5*block[4];} /*//dpfor (int i = 0; I <= block[3]; + + i) for (int j = 0; J <= Block[2]; + + j) Dp[i][j] = 100001;for (int i = 0; I &l T;= 4; + + i) for (int j = 0; J <= 9; + + j) Dp[i][j] = maps[i][j];for (int i = 0; I <= block[3]; + + i) for (int j = 0; J <= BLOCK[2]; + + j) {if (I >= 4 && J >= 0 && Dp[i][j] > dp[i-4][j-0]) dp[i][j] = dp[i-4][j-0]+1;if (i >= 3 && J >= 1 && DP I [j] > Dp[i-3][j-1]) dp[i][j] = dp[i-3][j-1]+1;if (i >= 2 && J >= 3 && dp[i][j] > dp[i-2][j-3]) d P[I][J] = dp[i-2][j-3]+1;if (i >= 1 && J >= 5 && Dp[i][j] > dp[i-1][j-5]) dp[i][j] = dp[i-1][j-5]+ 1;if (i >= 0 && J >= 9 && dp[i][j] > dp[i-0][j-9]) dp[i][j] = dp[i-0][j-9]+1;} DP *///greedy//3x3count + block[3]/4;block[3]%= 4;//2x2count + block[2]/9;block[2]%= 9;//greedy count + = Maps[block[3]][blo    Ck[2]];count + = (block[1]+9*block[3]+4*block[2]-36*maps[block[3]][block[2]]+35)/36;printf ("%d\n", count);} return 0;}

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UVa 311-packets