dfs-uva 10926 How Many Dependencies
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再次证明严谨的思维是ac的保证,这次自以为考虑到了所有情况,却栽在了数组这块,处理时从0开始,而输入依赖的工作序号却是从一开始,wrong了一下,找错找了半天才反应过来,果然是菜。
解题思路,简单dfs,看成一个有向图,不知道怎么描述,反正就是直接找每个点"依赖"的点有多少,输出最多的。
附代码
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int N;int rt[300][300],used[300];int ans;void dfs(int x){ for(int i=1;i<=N;i++){ if(rt[x][i]==1 && !used[i]){ used[i]=1; ans++; dfs(i); } }}int main(){ int num[300]; while(~scanf("%d",&N) && N!=0){ memset(rt,0,sizeof(rt)); for(int M=1;M<=N;M++){ int T; scanf("%d",&T); int yl; for(int i=0;i<T;i++){ scanf("%d",&yl); rt[M][yl]=1; } } memset(num,0,sizeof(num)); for(int i=1;i<=N;i++){ memset(used,0,sizeof(used)); for(int j=1;j<=N;j++){ ans=0; if(rt[i][j]==1 && !used[j]){ used[j]=1; ans++; dfs(j); num[i]+=ans; } } } int maxi=0; for(int i=1;i<=N;i++){ if(num[i]>num[maxi]){ maxi=i; } } printf("%d\n",maxi); } return 0;}
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