【CF】DZY Loves Chessboard
2014-07-15 16:02
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传送门:戳我
大致题意:给一个由'.'和'-'组成的矩阵,将'.'替换成W或B,使得W不和W相邻,B不和B相邻,输出任意一种方案,special judge。
思路:1、可以搜,搜出来一种就可以停下了;2、直接生成一个WBWBWB的100*100的矩阵,如果是'.',则直接输出对应位置的字母,保证不冲突
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int n,m,map[100+5][100+5];
int main(){
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++){
int cnt=i%2;
for (int j=1;j<=m;j++){
map[i][j]=cnt%2;
cnt++;
}
}
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++){
char c;
c=getchar();
while (c!='.' && c!='-') c=getchar();
if (c=='-') printf("-");
else {
if (map[i][j]==1) printf("B");
else printf("W");
}
}
if (i!=n) printf("\n");
}
return 0;
}
大致题意:给一个由'.'和'-'组成的矩阵,将'.'替换成W或B,使得W不和W相邻,B不和B相邻,输出任意一种方案,special judge。
思路:1、可以搜,搜出来一种就可以停下了;2、直接生成一个WBWBWB的100*100的矩阵,如果是'.',则直接输出对应位置的字母,保证不冲突
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int n,m,map[100+5][100+5];
int main(){
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++){
int cnt=i%2;
for (int j=1;j<=m;j++){
map[i][j]=cnt%2;
cnt++;
}
}
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++){
char c;
c=getchar();
while (c!='.' && c!='-') c=getchar();
if (c=='-') printf("-");
else {
if (map[i][j]==1) printf("B");
else printf("W");
}
}
if (i!=n) printf("\n");
}
return 0;
}
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