關於鄰接矩陣和鄰接表

關於臨界矩陣和鄰接表的使用情況

在很多圖論的題目中，如果題目本身沒有明確指出兩個頂點只有最多隻有一條直達的邊的話那麼最好使用鄰接表來存儲所有的邊，因爲鄰接矩陣會用某一條邊來覆蓋之前的邊，這條邊可以是最後輸入的邊，也可以是你設定了相應條件的邊，如果這下邊的邊權不一樣，就會很容易出現錯誤。比如求最短路的時候用之後出現的邊覆蓋了最短的邊，或者是求邊的入度出度的時候只記錄了一條邊以至於出錯。

HDU 3342 Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11758    Accepted Submission(s): 5532

Problem Description

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.

 

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.

 

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

 

Sample Input

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

 

Sample Output

YES
NO

使用鄰接矩陣的錯誤做法：

#include <iostream>
#include <queue>
#include <stdlib.h>
#include <stdio.h>
#include <map>
#include <string>
#include <cstdlib>
#include <stack>
#include <vector>
#include <math.h>
#include <algorithm>
#include <typeinfo>
#include <cstring>


using namespace std;
typedef long long ll;

const ll inf = 1061109567;
const int maxn = 1002;

int n,m;

int mat[maxn][maxn];

int vex_in[maxn];


int TopoSort(){
    queue<int> q;
    int s[maxn]={0};
    int cnt=0;
    for(int i=0;i<n;i++){
        if(vex_in[i]==0)    q.push(i);
    }

    while(!q.empty()){
        int p=q.front();
        q.pop();
        s[p]=1;
        cnt++;
        for(int i=0;i<n;i++){
            if(mat[p][i]){
                vex_in[i]--;
                if(vex_in[i]==0)    q.push(i);
            }
        }
    }
    /*for(int i=0;i<n;i++)
        if(!s[i])    return 0;*/
    if(cnt!=n)    return 0;
    return 1;
}

int main(int argc, char const *argv[])
{
    while(cin>>n>>m&&m&&n){
        for(int i=0;i<n;i++){
            vex_in[i]=0;
            for(int j=0;j<n;j++){
                mat[i][j]=0;
            }
        }

        int a,b;
        for(int i=0;i<m;i++){
            cin>>a>>b;
            mat[a][b]=1;
            vex_in[b]++;
        }
        if(TopoSort())
            cout<<"YES";
        else cout<<"NO";
        cout<<endl;
    }
    return 0;
}

此時莫名奇妙的wa，雖然我沒見到測試樣例，但是猜測可能是存在上述說的情況。於是改用鄰接表a了。

#include <iostream>
#include <queue>
#include <stdlib.h>
#include <stdio.h>
#include <map>
#include <string>
#include <cstdlib>
#include <stack>
#include <vector>
#include <math.h>
#include <algorithm>
#include <typeinfo>
#include <cstring>


using namespace std;
typedef long long ll;

const ll inf = 1061109567;
const int maxn = 1002;

int n,m;

int mat[maxn][maxn];

std::vector<int> edge[maxn];

int vex_in[maxn];


int TopoSort(){
	queue<int> q;
	//int s[maxn]={0};
	int cnt=0;
	for(int i=0;i<n;i++){
		if(vex_in[i]==0)	q.push(i);
	}

	while(!q.empty()){
		int p=q.front();
		q.pop();
		//s[p]=1;
		cnt++;
		int len=edge[p].size();
		for(int i=0;i<len;i++){
			vex_in[edge[p][i]]--;
			if(vex_in[edge[p][i]]==0)	q.push(edge[p][i]);
		}
	}
	/*for(int i=0;i<n;i++)
		if(!s[i])	return 0;*/
	if(cnt!=n)	return 0;
	return 1;
}

int main(int argc, char const *argv[])
{
	while(cin>>n>>m&&m&&n){
		for(int i=0;i<n;i++){
			vex_in[i]=0;
			edge[i].clear();
		}

		int a,b;
		for(int i=0;i<m;i++){
			cin>>a>>b;
			edge[a].push_back(b);
			vex_in[b]++;
		}
		if(TopoSort())
			cout<<"YES";
		else cout<<"NO";
		cout<<endl;
	}
	return 0;
}