# 2013-5-22 完美世界复赛第三场

A题,死了命的提示结果错误....  显然已模拟题.封装 remove 与 maintain 然后print.. 各种情况都考虑,将其后台数据都输出比较.还是未找到错误点在哪里..

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int N = 50;
char mp[N][N];
int n, m;
bool vis[N][N];

bool legal(int x,int y){
if(x>=1&&x<=n&&y>=1&&y<=m)
return true;
return false;
}
void maintain(){

for(int c = 1; c <= m; c++){

for(int r = n; r > 1; r--){
if( mp[r][c] == '0' ){
int p = -1;
for(int t = r-1; t >= 1; t-- ){
if( mp[t][c] != '0' ){
p = t; break;
}
}
if( p == -1 ) break;
else swap( mp[r][c], mp[p][c] );
}
}
}
}

bool find(int x,int y){
bool flag = false;
// left
if( y >= 3 ){
char ch = mp[x][y];
if( (mp[x][y-1]==ch) && (mp[x][y-2]==ch) ){
flag = true;
for(int i = y; i >= 1; i--)
if( mp[x][i] == ch ) vis[x][i] = true;
else break;
}
}
// right
if( y+2 <= m ){
char ch = mp[x][y];
if( (mp[x][y+1]==ch) && (mp[x][y+2]==ch) ){
flag = true;
for(int i = y; i <= m; i++)
if( mp[x][i] == ch ) vis[x][i] = true;
else break;
}
}
// up
if( x >= 3 ){
char ch = mp[x][y];
if( (mp[x-1][y]==ch) && (mp[x-2][y]==ch) ){
flag = true;
for(int i = x; i >= 1; i--)
if( mp[i][y] == ch ) vis[i][y] = true;
else break;
}
}
// down
if( x+2 <= n ){
char ch = mp[x][y];
if( (mp[x+1][y]==ch) && (mp[x+2][y]==ch) ){
flag = true;
for(int i = x; x <= n; i++)
if( mp[i][y] == ch ) vis[i][y] = true;
else    break;
}
}
return flag;
}
bool remove(){
memset(vis,0,sizeof(vis));
bool flag = false;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++){
if( (!vis[i][j]) && (mp[i][j]!='0') ){
if( find(i,j) )
flag = true;
}
}
if( flag ){
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
if( vis[i][j] ) mp[i][j] = '0';
}
return flag;
}
void print(){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++)
printf("%c", mp[i][j]);
puts("");
}
}

int main(){
freopen("1.in","r",stdin);
int T;
scanf("%d", &T);
while(T--){
scanf("%d%d",&n,&m);
for(int i = 1; i <= n; i++)
scanf("%s", mp[i]+1 );

maintain();
while( remove() ){
//    print();
maintain();
print();
}
}
return 0;
}
View Code

B题 . 魔方构造. 坑爹的是竟然还有N=2....  神一般的构造..看文库 http://wenku.baidu.com/view/02ace38ad0d233d4b14e693f.html   一般人都会..没看估计都不会.....

对于题目中 "对于每个N字图，每行输出N字图的一行，每行中的数字之间用一个或多个空格分开（注意对齐方式需要按最大的那个数字来对齐）", 真心想吐嘈... 这尼码是啥要求啊.所谓最大数字来对齐, 第一 是左还是右啊, 是 按最大的 n*n来, 还是当前一行的最大呢..  您好歹给点详细说明呗. 对于N=2这种无解的情况也没个说法,该如何处理.....

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int N = 110;

void gao(int mp[N][N], int n, int c){
int x = 0, y = n/2;
mp[x][y] = 1 + c;
for(int top = 2; top <= n*n; top++){
int x1 = (x-1+n)%n, y1 = (y+1+n)%n;
if( mp[x1][y1] )
x = (x+1+n)%n;
else x = x1, y = y1;
mp[x][y] = top + c;
}

}
void kao(int mp[N][N], int n){
if( n%4 == 0 ){
for(int i = 0, top = 1; i < n; i++)
for(int j = 0; j < n; j++)
mp[i][j] = top++;
for(int i = 0; i < n; i++){
for(int j = 0; j < n-i; j++){
if( (i==j)||(i+j==n-1) ) continue;
swap( mp[i][j], mp[n-1-i][n-1-j] );
}
}
}
}
int tmp[N][N];
int A[N][N], B[N][N], C[N][N], D[N][N];
int mp[N][N];

void cao(int n){
int m = n/4;
int k = 2*m+1;
gao( A, k, 0 ); gao( B, k, k*k );
gao( C, k, 2*k*k ); gao( D, k, 3*k*k );

for(int i = 0; i < k; i++){
int d = (i==k/2)?1:0;
for(int j = 0; j < m; j++){
swap( A[i][j+d], D[i][j+d] );
}
}
for(int i = 0; i < k; i++){
for(int j = 0; j < m-1; j++){
swap( C[i][k-1-j], B[i][k-1-j] );
}
}
for(int i = 0; i < k; i++)
for(int j = 0; j < k; j++){
mp[i][j] = A[i][j];
mp[i][j+3] = C[i][j];
mp[i+3][j] = D[i][j];
mp[i+3][j+3] = B[i][j];

}
}
int n;

void print(int x){
int la = 0, lb = 0, t = n*n;
while( t ) la++, t /= 10;
t = x;
while( t ) lb++, t /= 10;
printf("%d", x);
for(int i = lb; i < la; i++)
printf(" ");
}
int main(){
while( scanf("%d", &n), n ){
memset( mp, 0, sizeof(mp));

if( n == 2 ){ continue; }
if( n&1 ) gao(mp, n, 0);
else if( n%4 == 0 ) kao(mp, n);
else cao( n );
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if( j != 0 ) printf(" ");
//printf("%d", mp[i][j] );
print( mp[i][j] );
}
puts("");
}
puts("");
}
return 0;
}
View Code

C题, 不知道啥题..不过 岛娘A掉了.. Orz...

posted @ 2013-05-22 23:03  yefeng1627  阅读(233)  评论(0编辑  收藏  举报