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|

+-+

---

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;}