poj 1129 DFS 无剪枝就可AC

 这道题似乎也没什么剪枝可写，没有细想。

就是DFS。

还是，写代码的过程中牢牢记住DFS的思想即可。

代码如下：

#include <iostream>#include <string>using namespace std;int rep[26][27],channelChosen[26];bool check(int curRep,int curCha,int amount){for (int j = 1;j <= rep[curRep][0];j++){if(channelChosen[curRep]==channelChosen[rep[curRep][j]-1])return false;}return true;}int DFS(int curRepeater,int curChannel,int num){if(curRepeater>=num)return curChannel;for (int i = 1;i <= curChannel;i++){channelChosen[curRepeater]=i;if(check(curRepeater,curChannel,num)){return DFS(curRepeater+1,curChannel,num);}elsecontinue;}channelChosen[curRepeater]=curChannel+1;return DFS(curRepeater+1,curChannel+1,num);}int main(){int numOfRepeaters;while (cin>>numOfRepeaters&&numOfRepeaters!=0){string str;for (int i = 0;i < numOfRepeaters;i++){cin>>str;int j = 1;string::iterator it = str.begin()+2;for (;it != str.end();it++){rep[i][j++] = *it-'A'+1;}rep[i][0] = j-1;}int result = DFS(0,1,numOfRepeaters);string res = (result==1)?" channel needed.":" channels needed.";cout<<result<<res<<endl;}}