hdu2614 Beat BFS or DFS

Beat

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1286    Accepted Submission(s): 751

Problem Description

Zty is a man that always full of enthusiasm. He wants to solve every kind of difficulty ACM problem in the world. And he has a habit that he does not like to solve
a problem that is easy than problem he had solved. Now yifenfei give him n difficulty problems, and tell him their relative time to solve it after solving the other one.
You should help zty to find a order of solving problems to solve more difficulty problem. 
You may sure zty first solve the problem 0 by costing 0 minute. Zty always choose cost more or equal time’s problem to solve.

 

Input

The input contains multiple test cases.
Each test case include, first one integer n ( 2< n < 15).express the number of problem.
Than n lines, each line include n integer Tij ( 0<=Tij<10), the i’s row and j’s col integer Tij express after solving the problem i, will cost Tij minute to solve the problem j.

 

Output

For each test case output the maximum number of problem zty can solved.

 

Sample Input

   
   
   
   

    
    
    
    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
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    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. 
    
   
   
   
   

BFS

//hdu2614 Beat BFS
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <iostream>

using namespace std;

const int maxn = 15 + 10;

struct node {
	//pos是当前位置，maxx是到目前为止最大的Tij值，cnt计数
	int pos, maxx, cnt;
	//m用来标记是否走过，m[i] > 0走过，m[i] == 0没有
	//每个节点相当于一个状态，都要有自己的标记情况
	map<int, int> m;
	node() {}
	node(int pos, int maxx, int cnt) : pos(pos), maxx(maxx), cnt(cnt) {}
};

int n, mapp[maxn][maxn], ans;

void bfs() {
	queue<node> Q;

	node ns = node(0, 0, 1);
	ns.m[0]++;

	Q.push(ns);

	while (!Q.empty()) {
		node tn = Q.front(); Q.pop();
		ans = max(ans, tn.cnt);
		int p = tn.pos, maxx = tn.maxx;
		for (int i = 0; i < n; i++) {
			//自己不能到自己，不重复走，不走更小的
			if (i != p && tn.m[i] == 0 && mapp[p][i] >= maxx) {
				node in = tn;
				in.pos = i; in.cnt++; in.maxx = mapp[p][i];
				in.m[i]++;
				Q.push(in);
			}
		}
	}
}

int main()
{
	while (~scanf("%d", &n)) {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				scanf("%d", &mapp[i][j]);
			}
		}
		ans = -1;
		bfs();
		printf("%d\n", ans);
	}
	return 0;
}

DFS

//hdu2614 Beat DFS
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 15 + 10;
int mapp[maxn][maxn], ans, n;
bool vis[maxn];

//解到第dep个题，此时最大的时间值为maxx，cnt为已经将dep计算在内的题目数量
void dfs(int dep, int maxx, int cnt) {
	ans = max(ans, cnt);
	for (int i = 0; i < n; i++) {
		//去重，不走自己，不走更小的
		if (!vis[i] && dep != i && mapp[dep][i] >= maxx) {
			vis[i] = true;
			dfs(i, mapp[dep][i], cnt + 1);
			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", &mapp[i][j]);
			}
		}
		ans = -1;
		memset(vis, false, sizeof(vis));
		vis[0] = true;
		dfs(0, 0, 1);
		vis[0] = false;
		printf("%d\n", ans);
	}
	return 0;
}