TOJ 3365 ZOJ 3232 It's not Floyd Algorithm / 强连通分量

It's not Floyd Algorithm

时间限制(普通/Java):1000MS/3000MS     运行内存限制:65536KByte
 

描述

 

When a directed graph is given, we can solve its transitive closure easily using the well-known Floyd algorithm.

But if you're given a transitive closure, can you give a corresponding directed graph with minimal edges?

输入

 

About 100 test cases, seperated by blank line.

First line of each case is an integer N (1<=N<=200). The followingN lines represent the given transitive closure in 0-1 matrix form, each line hasN numbers.

输出

 

For each case, just output the number of minimal edges of a directed graph which has a given transitive closure.

样例输入

 

1
1

2
1 0
0 1

2
1 1
1 1

3
1 1 1
0 1 1
0 0 1

样例输出

 

0
0
2
2

提示

Transitive closure can be presented as a matrix T, where T_i,j is true if and only if there is a path from vertexi toj.

首先缩点 在建图 对于每个强连通分量如果有n个点那么最少只需n条边就可以联通（n = 1 除外）

然后对于缩点后的图咋做一次反闭包 去掉多余的边 在统计一下

	
#include<stdio.h>
#include<string.h>
int n;
int dfn[210];
int low[210];
bool instack[210];
int stack[210];
int cnt,num,top;
int a[210][210];
int count[210];
int belong[210];
int map[210][210];
void floyd()
{
	int i,j,k;
	for(k = 1;k <= cnt; k++)
		for(i = 1;i <= cnt; i++)
			for(j = 1;j <= cnt; j++)
				if(map[i][k] && map[k][j] && map[i][j])
					map[i][j] = 0;
}
void tarjan(int i)
{
        int j,k;
        dfn[i] = low[i] = ++num;
        instack[i] = true;
        stack[++top] = i;
        for(j = 1;j <= n; j++)
        {
            k = a[i][j];
            if(!k)
            	continue;
            if(!dfn[j])
            {
                tarjan(j);
                if(low[i] > low[j])
                	low[i] = low[j];
            }
            else if(instack[j] && low[i] > dfn[j])
                low[i] = dfn[j];
        } 
        if(low[i] == dfn[i])
        {
            cnt++;
            do
            {
                j = stack[top--];
                instack[j] = false;
                belong[j] = cnt; 
                count[cnt]++;
            }
            while(i != j);
        }
} 
int main()
{
	int i,j,sum;
	while(scanf("%d",&n)!=EOF)
	{
		for(i = 1;i <= n; i++)
			for(j = 1; j<= n; j++)
				scanf("%d",&a[i][j]);
		num = top = cnt = 0;
		memset(dfn,0,sizeof(dfn));
		memset(instack,false,sizeof(instack));
		memset(count,0,sizeof(count));
		memset(map,0,sizeof(map));
		for(i = 1;i <= n; i++)
			if(!dfn[i])
				tarjan(i);
		for(i = 1; i<= n; i++)
			for(j = 1; j<= n; j++)
				if(a[i][j])
					if(belong[i] != belong[j])
						map[belong[i]][belong[j]] = 1;
		sum = 0;
		floyd();
		for(i = 1;i <= cnt; i++)
			if(count[i] != 1)
				sum += count[i];
		for(i = 1;i <= cnt; i++)
			for(j = 1; j<= cnt; j++)
				if(map[i][j])
					sum++;
		printf("%d
",sum);
	}
	return 0;
}