Codeforces Round #254(div2)A
2014-08-19 02:26
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很有趣的题。想到了就非常简单,想不到就麻烦了。
其实就是一种逆向思维:最后结果肯定是这样子:
WBWBWBWB...
BWBWBWBW...
WBWBWBWB...
...
里面有“-”的地方改成“-”就行了。
但是我开始是正着想的,想每个点怎么处理,这还要看它周围点的状态,越想越麻烦。。。
这题中体现的正难则反的逆向思维很值得学习。
其实就是一种逆向思维:最后结果肯定是这样子:
WBWBWBWB...
BWBWBWBW...
WBWBWBWB...
...
里面有“-”的地方改成“-”就行了。
但是我开始是正着想的,想每个点怎么处理,这还要看它周围点的状态,越想越麻烦。。。
这题中体现的正难则反的逆向思维很值得学习。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<stack> #include<queue> using namespace std; #define INF 1000000000 #define eps 1e-8 #define pii pair<int,int> #define LL long long int int n,m; char mp[105][105]; int main() { //freopen("in1.txt","r",stdin); //freopen("out.txt","w",stdout); scanf("%d%d",&n,&m); getchar(); for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) { scanf("%c",&mp[i][j]); } getchar(); } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(mp[i][j]=='-') { printf("-"); } else { if((i+j)&1) { printf("B"); } else printf("W"); } } printf("\n"); } //fclose(stdin); //fclose(stdout); return 0; }
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