uva 11902 - Dominator

D

Dominator

In graph theory, a node X dominates a node Y if every path from the predefined start node to Y must go through X. If Y is not reachable from the start node then node Y does not have any dominator. By definition, every node reachable from the start node dominates itself. In this problem, you will be given a directed graph and you have to find the dominators of every node where the 0^th node is the start node.

As an example, for the graph shown right, 3 dominates 4 since all the paths from 0 to 4 must pass through3. 1 doesn't dominate 3 since there is a path 0-2-3 that doesn't include 1.

Input

The first line of input will contain T (≤ 100) denoting the number of cases.

Each case starts with an integer N (0 < N < 100) that represents the number of nodes in the graph. The next N lines contain N integers each. If the j^th(0 based) integer of i^th(0 based) line is 1, it means that there is an edge from node i to node j and similarly a 0 means there is no edge.

Output

For each case, output the case number first. Then output 2N+1 lines that summarizes the dominator relationship between every pair of nodes. If node A dominates node B, output 'Y' in cell (A, B), otherwise output 'N'. Cell (A, B) means cell at A^th row and B^th column. Surround the output with |, + and – to make it more legible. Look at the samples for exact format.

Sample Input	Output for Sample Input
2 5 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1	Case 1: +---------+ |Y|Y|Y|Y|Y| +---------+ |N|Y|N|N|N| +---------+ |N|N|Y|N|N| +---------+ |N|N|N|Y|Y| +---------+ |N|N|N|N|Y| +---------+ Case 2: +-+ |Y| +-+

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Problem Setter: Sohel Hafiz, Special Thanks: Kazi Rakibul Hossain, Jane Alam Jan

n的范围很小，枚举每个点，去掉它相邻的边，然后dfs找从0能到达的所有点，复杂度O(n^3)。需要注意的是 If Y is not reachable from the start node then node Y does not have any dominator.没特判这个wa了一次，然后因为可能有环，dfs时注意走过的点就不要再搜下去了，re一次。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 100 + 5;

int n;
int G[maxn][maxn];
int temG[maxn][maxn];
int vis[maxn];
int ans[maxn][maxn];

void dfs(int x){
    vis[x] = 1;
    for(int i = 0;i < n;i++){
        if(temG[x][i] == 1 && vis[i] == 0)
            dfs(i);
    }
}

void print(){
    cout << '+';
    for(int i = 0;i < 2*n-1;i++) cout << '-';
    cout << '+' << endl;
}

int main(){
    int t, kase = 0;
    scanf("%d", &t);
    while(t--){
        kase++;
        scanf("%d", &n);
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                scanf("%d", &G[i][j]);
            }
        }

        memset(ans, 0, sizeof(ans));
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                for(int k = 0;k < n;k++){
                    if(j == i || k == i) temG[j][k] = 0;
                    else temG[j][k] = G[j][k];
                }
            }
            memset(vis, 0, sizeof(vis));
            dfs(0);
            for(int j = 0;j < n;j++){
                if(vis[j] == 1){
                    ans[i][j] = 1;
                }
            }
        }
        ans[0][0] = 0;

        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++)
                temG[i][j] = G[i][j];
        }
        memset(vis, 0, sizeof(vis));
        dfs(0);
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                if(vis[i] == 0 || vis[j] == 0){
                    ans[i][j] = 1;
                }
            }
        }

        printf("Case %d:\n", kase);
        print();
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                cout << '|';
                if(ans[i][j] == 0) cout << 'Y';
                else cout << 'N';
            }
            cout << '|' << endl;
            print();
        }
    }
    return 0;
}