1103:Ancient Messages
数一数就能发现,题目表中的6个符号从左到右依次有1,3,5,4,0,2个洞,各不相同。这样,只需要数一数输入的符号有几个“白洞”,就能准确地知道它是哪个符号了。
至于具体实现,我的思路是:先确定一个符号,找到该符号的空洞个数然后将该符号“擦除”,继续确定下一个符号。具体而言,每次找到一个黑点,然后DFS,如果其周围点为黑点则递归,若为白点且为封闭块则个数加1,最后再次DFS将该符号的所有黑点标记为-1。期间犯了两个低级错误(大小写写反、变量名写错)找了半天bug。。。自己电脑上测试的样例其中一个应该是栈溢出了,但是提交上去还是AC了。
version 1(20ms):
#include<bits/stdc++.h>
using namespace std;
const int maxh = 200 + 5;
const int maxw = 50 + 5;
int m,n,cnt;
int cnts[6];
int graph[maxh][maxw*4];
char symbol[6] = { 'A', 'D', 'J', 'K', 'S', 'W' };
int seq[6] = {1,5,3,2,4,0};
void convert(int i,int j){ //进制转换
char c = getchar();
int num = (c <= '9' ? c - '0' : c - 'a' + 10); //小写字母!!!
j *= 4;
int t = 0;
while(1<<t <= num) t++;
for(int k = 0;k < (4-t);k++) graph[i][j+k] = 0;
j += 4-t;
for(;t;t--){
if(num >= 1<<(t-1)){ num -= 1<<(t-1); graph[i][j++] = 1; }
else graph[i][j++] = 0;
}
}
int dfs1(int x,int y){ //判断该连通块是否被黑色像素包围
int mark = 1;
if(!x || !y || x == m-1 || y == n-1) mark = 0; //一定要判断自身是否在边界 如下:
// 1 1
// 1 0
// 右下角这个点如果不首先判断自身的话就会被误判为封闭块
graph[x][y] = -1;
for(int dx = -1;dx <= 1;dx++){
for(int dy = -1;dy <= 1;dy++){
int nx = x + dx,ny = y + dy; //be careful!! don't change x and y
if((!dx && !dy) || nx >= m || nx < 0 || ny >= n || ny < 0
|| graph[nx][ny] != 0) continue;
if(!nx || !ny || nx == m-1 || ny == n-1) mark = 0; //若能到边界 则非封闭
if(!dfs1(nx,ny)) mark = 0;
}
}
return mark;
}
void dfs2(int x,int y){ //寻找黑色点
graph[x][y] = 2;
for(int dx = -1;dx <= 1;dx++){
for(int dy = -1;dy <= 1;dy++){
int nx = x + dx,ny = y + dy;
if((!dx && !dy) || nx >= m || nx < 0 || ny >= n || ny < 0
|| graph[nx][ny] == 2 || graph[nx][ny] == -1) continue; //nx写成x找了半天bug。。
if(graph[nx][ny] == 1) dfs2(nx,ny);
else cnt += dfs1(nx,ny);
}
}
}
void dfs3(int x,int y){ //将该符号的所有像素擦除
graph[x][y] = -1;
for(int dx = -1;dx <= 1;dx++){
for(int dy = -1;dy <= 1;dy++){
int nx = x + dx,ny = y + dy;
if((!dx && !dy) || nx >= m || nx < 0 || ny >= n || ny < 0
|| graph[nx][ny] != 2) continue;
dfs3(nx,ny);
}
}
}
int main(){
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
int kase = 0;
while(scanf("%d %d",&m,&n) == 2 && m){
for(int i = 0;i < m;i++){
getchar();
for(int j = 0;j < n;j++) convert(i,j);
}
n *= 4;
memset(cnts,0,sizeof(cnts));
for(int i = 0;i < m;i++){
for(int j = 0;j < n;j++){
if(graph[i][j] == 1){
cnt = 0;
dfs2(i,j);
cnts[cnt]++;
dfs3(i,j);
}
}
}
printf("Case %d: ",++kase);
for(int i = 0;i < 6;i++){
while(cnts[seq[i]]--) putchar(symbol[i]);
}
putchar('\n');
}
return 0;
}
其实是四连通,检查四个方向就够了,进制转换利用位运算更方便,还有就是之前想多了,直接将黑点直接标记为-1即可,不需要最后再DFS一次:
version 2(10ms):
#include<bits/stdc++.h>
using namespace std;
const int maxh = 200 + 5;
const int maxw = 50 + 5;
int m,n,cnt;
int cnts[6];
int graph[maxh][maxw*4];
char symbol[7] = "ADJKSW";
int seq[6] = {1,5,3,2,4,0};
void convert(int i,int j){ //进制转换
char c = getchar();
int num = (c <= '9' ? c - '0' : c - 'a' + 10); //小写字母!!!
j *= 4;
for(int k = 3;k >= 0;k--){
graph[i][j++] = 1 & (num>>k);
}
}
int dfs1(int x,int y){ //判断该连通块是否被黑色像素包围
int mark = 1;
if(!x || !y || x == m-1 || y == n-1) mark = 0; //一定要判断自身是否在边界 如下:
// 1 1
// 1 0
// 右下角这个点如果不首先判断自身的话就会被误判为封闭块
graph[x][y] = -1;
for(int dx = -1;dx <= 1;dx++){
for(int dy = -1;dy <= 1;dy++){
int nx = x + dx,ny = y + dy; //be careful!! don't change x and y
if(dx == dy || (!dx && !dy) || nx >= m || nx < 0 || ny >= n || ny < 0
|| graph[nx][ny] != 0) continue;
if(!nx || !ny || nx == m-1 || ny == n-1) mark = 0; //若能到边界 则非封闭
if(!dfs1(nx,ny)) mark = 0;
}
}
return mark;
}
void dfs2(int x,int y){ //寻找黑色点
graph[x][y] = -1;
for(int dx = -1;dx <= 1;dx++){
for(int dy = -1;dy <= 1;dy++){
int nx = x + dx,ny = y + dy;
if(dx == dy || (!dx && !dy) || nx >= m || nx < 0 || ny >= n || ny < 0
||graph[nx][ny] == -1) continue; //nx写成x找了半天bug。。
if(graph[nx][ny] == 1) dfs2(nx,ny);
else cnt += dfs1(nx,ny);
}
}
}
int main(){
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
int kase = 0;
while(scanf("%d %d",&m,&n) == 2 && m){
for(int i = 0;i < m;i++){
getchar();
for(int j = 0;j < n;j++) convert(i,j);
}
n *= 4;
memset(cnts,0,sizeof(cnts));
for(int i = 0;i < m;i++){
for(int j = 0;j < n;j++){
if(graph[i][j] == 1){
cnt = 0;
dfs2(i,j);
cnts[cnt]++;
}
}
}
printf("Case %d: ",++kase);
for(int i = 0;i < 6;i++){
while(cnts[seq[i]]--) putchar(symbol[i]);
}
putchar('\n');
}
return 0;
}
但是还有一份因为栈溢出没过的样例还没找到原因,先放这儿吧,放一份这位博主的代码,有很多值得借鉴的地方:
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
const int maxn=210;
const int a[6]={1,5,3,2,4,0};
const char b[7]="WAKJSD";
int h,w,cnt[6],ct;
bool vis[maxn][maxn],g[maxn][maxn];
map<char,int>hex;
void pre(){
for(int i=0;i<10;i++) hex['0'+i]=i;
for(int i=0;i<6;i++) hex['a'+i]=10+i;
return;
}
bool read()
{
scanf("%d%d",&h,&w);
if(!h&&!w) return false;
memset(g,0,sizeof(g));
for(int i=1;i<=h;i++){
getchar();
int pos=1;
for(int j=1;j<=w;++j){
int c=hex[getchar()];
for(int k=3;k>=0;--k)
g[i][pos++]=c&(1<<k);
}
}
w*=4;
return true;
}
bool in(int x,int y){
return x>=0&&x<=h+1&&y>=0&&y<=w+1;
}
void dfs(int x,int y,bool k){
if(!in(x,y)||vis[x][y]) return;
if(!k&&g[x][y]) return;
if(k&&!g[x][y]){
++ct;
dfs(x,y,0);
return;
}
vis[x][y]=true;
dfs(x,y+1,k);
dfs(x+1,y,k);
dfs(x,y-1,k);
dfs(x-1,y,k);
return;
}
void solve()
{
memset(cnt,0,sizeof(cnt));
memset(vis,0,sizeof(vis));
dfs(0,0,0);
for(int i=0;i<=h;i++){
for(int j=0;j<=w;j++){
if(g[i][j]&&!vis[i][j]){
ct=0;
dfs(i,j,1);
++cnt[ct];
}
}
}
for(int i=0;i<6;i++)
while(cnt[a[i]]--) putchar(b[a[i]]);
printf("\n");
return;
}
int main()
{
pre();
int t=0;
while(read()){
printf("Case %d: ",++t);
solve();
}
return 0;
}
Email:JingwangLi@outlook.com