标签:col print mes namespace oid als stream har div
白书第一章例题6
构造。思维。几何。
分别从几个角度去看,有矛盾就删掉,最后遍历一下统计个数
方法证明:第一个方块肯定要删除。假设前k个必须删除,第k+1个矛盾出现,假如不删掉,矛盾将持续存在,故必须删掉。
代码有很多细节。
比如注意宏定义加() //#define REP(i,n) for(int i=0;i<(n);i++)
#include<iostream> using namespace std; #define REP(i,n) for(int i=0;i<(n);i++) int n; char read_char() { char c; for (;; ) { c = getchar(); if ((c >= ‘A‘&&c <= ‘Z‘) || c == ‘.‘) return c; } } void get(int k, int i, int j, int p, int &x, int &y, int &z) { if (k == 0) { x = p; y = j; z = i; } if (k == 1) { x = n - 1 - j; y = p; z = i; } if (k == 2) { x = n - 1 - p; y = n - 1 - j; z = i; } if (k == 3) { x = j; y = n - p - 1; z = i; } if (k == 4) { x = n - 1 - i; y = j; z = p; } if (k == 5) { x = i; y = j; z = n - 1 - p; } } ; char view[6][10][10], pos[10][10][10]; int main() { while (cin >> n) { if (n == 0)break; REP(i, n) REP(k, 6) REP(j, n) view[k][i][j] = read_char(); REP(i, n) REP(j, n) REP(k, n) pos[i][j][k] = ‘#‘; REP(k, 6) REP(i, n) REP(j, n) if (view[k][i][j] == ‘.‘) { REP(p, n) { int x, y, z; get(k, i, j, p, x, y, z); pos[x][y][z] = ‘.‘; } }; for (;;) { bool done = true; REP(i, n)REP(j, n)REP(k, 6) if (view[k][i][j]!=‘.‘ ) { REP(p, n) { int x, y, z; get(k, i, j, p, x, y, z); if (pos[x][y][z] == ‘.‘) continue; if (pos[x][y][z] == ‘#‘) { pos[x][y][z] = view[k][i][j]; break; } if (pos[x][y][z] == view[k][i][j]) break; pos[x][y][z] = ‘.‘; done = false; } } if (done) break; } int ans = 0; REP(i, n)REP(j, n)REP(k, n) if (pos[i][j][k] != ‘.‘) ans++; printf("Maximum weight: %d gram(s)\n", ans); } return 0; }
标签:col print mes namespace oid als stream har div
原文地址:https://www.cnblogs.com/lqerio/p/9860918.html