[UVA][圖論] 11550 - Demanding Dilemma
Problem D
DEMANDING DILEMMA
A simple undirected graph is an ordered pair G = (V, E) where V is a non-empty set of vertices, and E is a set of unordered pairs (u, v) where u and v are in V and u ≠ v. If S is a set, define |S| as the size of S. An incidence matrix M is a |V| x |E| matrix where M(i, j) is 1 if edge j is incident to vertex i (edge j is either (i, u) or (u, i)) and 0 otherwise.
Given an n x m matrix, can it be an incidence matrix of a simple undirected graph G = (V, E) where |V| = n and |E| = m?
Program Input
The first line of the input contains an integer t (1 ≤ t ≤ 41), the number of test cases.
Each test case starts with a line with two integers n (1 ≤ n ≤ 8) and m (0 ≤ m ≤ n(n-1)/2). Then n lines containing m integers (0’s or 1’s) each follow such that the jth number on the ith line is M(i, j).
Program Output
For each test case print “Yes” if the incidence matrix given in the input can be an incidence matrix of some simple undirected graph, otherwise print “No”.
Sample Input & Output
INPUT
3 3 3 1 1 0 0 1 1 1 0 1 3 1 1 1 0 3 3 1 1 0 1 1 1 1 0 0OUTPUT
Yes Yes No
Calgary Collegiate Programming Contest 2008
題目在描述離散數學中講的點邊記錄方式，
然後簡單它是否為一個簡單圖(無重邊)。
點邊記錄方式是這麼描述的的
------e1--e2--e3--e4--e5 ...
node1 0 0 1
node2 1 1 0
node3 1 0 1
node4 0 1 0 ...
表示這個點是否有在這邊的兩端點，因此邊不會同具有非兩個點的情況。
同時要檢驗是否有重邊。
#include <stdio.h>
int main() {
int vg[10][100];
int n, m, i, j;
int testcase;
scanf("%d", &testcase);
while(testcase--) {
scanf("%d %d", &n, &m);
for(i = 0; i < n; i++)
for(j = 0; j < m; j++)
scanf("%d", &vg[i][j]);
int g[10][10] = {}, err = 0, x, y;
for(i = 0; i < m && !err; i++) {
int cnt = 0;
for(j = 0; j < n; j++) {
if(vg[j][i]) {
if(cnt == 0) x = j;
else y = j;
cnt++;
}
}
if(cnt != 2) err = 1;
else {
if(g[x][y]) err = 1;
g[x][y] = g[y][x] = 1;
}
}
puts(err ? "No" : "Yes");
}
return 0;
}