HDU 2614 Beat (DFS)

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题意： n*n的图,a[i][j]:表示做完i题目在做j题目需要耗费的时间，每次选择的题目耗费时间要不小于之前的，问最多能做几题

#include<iostream>#include<cstdlib>#include<cmath>#include<algorithm>#include<cstring>#include<cstdio>#include<queue>#include<set>#include<stack>#define cl(a,b) memset(a,b,sizeof(a));#define LL long long#define P pair<int,int>#define X first#define Y second#define out(x) cout<<x<<endl;using namespace std;const int maxn=20;const int inf=9999999;int a[maxn][maxn];int n;bool vis[maxn];int ans;void dfs(int row,int sum,int time){    ans=max(ans,sum);    if(sum==n){        return;    }    for(int i=1;i<n;i++)if(!vis[i]){        if(a[row][i]>=time){            vis[i]=true;            dfs(i,sum+1,a[row][i]);            vis[i]=false;        }    }}int main(){    while(~scanf("%d",&n)){        for(int i=0;i<n;i++){            for(int j=0;j<n;j++){                scanf("%d",&a[i][j]);            }        }        cl(vis,false);        ans=0;        dfs(0,0,0);        printf("%d\n",ans+1);    }    return 0;}

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