HDU 2614 Beat (dfs)

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3
0 0 0
1 0 1
1 0 0
3
0 2 2
1 0 1
1 1 0
5
0 1 2 3 1
0 0 2 3 1
0 0 0 3 1
0 0 0 0 2
0 0 0 0 0

3
2
4
Hint
Hint: sample one, as we know zty always solve problem 0 by costing 0 minute. 
So after solving problem 0, he can choose problem 1 and problem 2, because T01 >=0 and T02>=0. 
But if zty chooses to solve problem 1, he can not solve problem 2, because T12 < T01. 
So zty can choose solve the problem 2 second, than solve the problem 1.  
 

题意：有n道题，a[i][j]表示做完i题后做j题的难度，这个人从第0题开始做，每次做不比上次题目简单的题，问他最多做多少题？

理解题意半天，解题还是比较顺利，解释在代码中

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;

#define N 20

int a[N][N],n;
int ans,vis[N];


void dfs(int le,int num,int s) //le表示刚做的题，例如le=0，那么现在在第0行选择做的下一题，s是刚做题的难度
{
    int i;

    if(num>ans) ans=num; //num是已经做的题

    for(i=0;i<n;i++)
    {
        if(vis[i]||a[le][i]<s) continue;//这个题做过，或者现在做i题难度没有s（前一题的分）高

        vis[i]=1;   //假设做i题
        dfs(i,num+1,a[le][i]);
        vis[i]=0;   //取消假设
    }
}

int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            scanf("%d",&a[i][j]);

        memset(vis,0,sizeof(vis));
        vis[0]=1; //题目说先做第0题
        ans=0;
        dfs(0,1,0);

        printf("%d\n",ans);
    }
    return 0;
}

HDU 2614 Beat (dfs),布布扣,bubuko.com

HDU 2614 Beat (dfs)

标签：des   style   color   os   io   for   ar   div   

原文地址：http://blog.csdn.net/u014737310/article/details/38586953