Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B
Claw Decomposition

Input: Standard Input

Output: Standard Output

 

A claw is defined as a pointed curved nail on the end of each toe in birds, some reptiles, and some mammals. However, if you are a graph theory enthusiast, you may understand the following special class of graph as shown in the following figure by the word claw.

If you are more concerned about graph theory terminology, you may want to define claw as K_1,3.

 

Let’s leave the definition for the moment & come to the problem. You are given a simple undirected graph in which every vertex has degree 3. You are to figure out whether the graph can be decomposed into claws or not.

 

Just for the sake of clarity, a decomposition of a graph is a list of subgraphs such that each edge appears in exactly one subgraph in the list.

 

Input

 

There will be several cases in the input file. Each case starts with the number of vertices in the graph, V (4<=V<=300). This is followed by a list of edges. Every line in the list has two integers, a & b, the endpoints of an edge (1<=a,b<=V). The edge list ends with a line with a pair of 0. The end of input is denoted by a case with V=0. This case should not be processed.

 

Output

 

For every case in the input, print YES if the graph can be decomposed into claws & NO otherwise.

 

Sample Input                                                  Output for Sample Input

4 1 2 1 3 1 4 2 3 2 4 3 4 0 0 6 1 2 1 3 1 6 2 3 2 5 3 4 4 5 4 6 5 6 0 0 0

NO NO

---

題目是要檢查爪子, 但是爪子可能重疊在一起, 我們必須分解能不能都是爪子,
如此一來可以判定, 只要是二分圖就能拆成很多爪子, 但是一個圖形可以是很多二分圖,
因此要注意這一點。

#include <stdio.h>
#include <vector>
#include <queue>
#include <string.h>
using namespace std;
bool checkBipartite(int V, vector<int> map[], bool color[], bool used[], int st) {
    queue<int> Q;
    Q.push(st);
    int tn;
    color[st] = true, used[st] = true;
    while(!Q.empty()) {
        tn = Q.front();
        Q.pop();
        for(vector<int>::iterator i = map[tn].begin(); i != map[tn].end(); i++) {
            if(used[*i] == false) {
                used[*i] = true;
                color[*i] = !color[tn];
                Q.push(*i);
            } else {
                if(color[*i] == color[tn])
                    return false;
            }
        }
    }
    return true;
}
int main() {
    int V, x, y;
    vector<int> map[301];
    while(scanf("%d", &V) == 1 && V) {
        for(int i = 0; i <= V; i++)
            map[i].clear();
        while(scanf("%d %d", &x, &y) == 2) {
            if(x == 0 && y == 0)
                break;
            map[x].push_back(y);
            map[y].push_back(x);
        }
        bool color[V+1], used[V+1], ans = true;
        memset(color, 0, sizeof(color));
        memset(used, 0, sizeof(used));
        for(int i = 1; i <= V; i++) {
            if(used[i] == false) {
                ans &= checkBipartite(V, map, color, used, i);
            }
        }
        if(ans == true)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}