UVA 434 Matty's Blocks

M比较好计算,N弄了半天时间.

首先对于front面的每一个高度front[i],在rightt面里面尽量先找相等的rightt[j],满足没有被其他相同高度front[k] (k != i)的匹配掉,如果找到了就填上front[i],如果没有相等的,找大于front[i]的最右边的rightt[j],然后填掉,如果依旧找不到,那么就留给rightt那边处理.

对于rightt做法相同.

#include <iostream>#include <cstdio>#include <memory.h>#include <algorithm>using namespace std;const int maxn = 9;int K;int sq[maxn][maxn], front[maxn], rightt[maxn];int used[maxn];int main(){int T;scanf("%d", &T);while(T--){scanf("%d", &K);memset(sq, 0, sizeof(sq));memset(used, 0, sizeof(used));for (int i = 0; i < K; ++i){scanf("%d", &front[i]);}for (int i = 0; i < K; ++i){scanf("%d", &rightt[i]);}int ansN = 0, ansM = 0;for (int i = 0; i < K; ++i){int idx = -1, ff = 0;for (int j = 0; j < K; ++j){if(rightt[j] == front[i]){idx = j;ff = 1;if(!used[j]){used[j] = 1;idx = j;break;}}else if(rightt[j] > front[i] && !ff){idx = j;}}if(idx != -1){sq[i][idx] = front[i];}}for (int i = 0; i < K; ++i){int fl = 0;for (int j = 0; j < K; ++j){if(sq[j][i] == rightt[i]){fl = 1; break;}}if(fl == 0){int idx = -1, ff = 0;for (int j = 0; j < K; ++j){if(front[j] == rightt[i]){idx = j;ff = 1;if(!used[j]){used[j] = 1;idx = j;break;}}else if(front[j] > rightt[i] && !ff){idx = j;break;}}if(idx != -1){sq[idx][i] = rightt[i];}}}for (int i = 0; i < K; ++i){for (int j = 0; j < K; ++j){ansN += sq[i][j];ansM += min(front[i], rightt[j]) - sq[i][j];}}printf("Matty needs at least %d blocks, and can add at most %d extra blocks.\n", ansN, ansM);}return 0;}