给定的阵列ARR [] N对代表一个有向图，其中每对字符串的表示由一对字符串的表示的节点之间的边缘字符串的，任务是检查给定的有向图的结构是否是和石头剪刀布游戏一样。如果发现为真，则打印“是”。否则，打印“否”。
笔记：
In the real game of Rock-Paper-Scissor, the rock dominates over the scissor, scissor dominates over the paper and paper dominates over the rock.The game can be modelled using a directed graph as below, in which the directed edge (a, b) shows that a dominates over b.
例子：
Input: arr[] = {{“Snake”, “Water”}, {“Water”, “Gun”}, {“Gun”, “Snake”}}
Output: Yes
Explanation:
Therefore, from the above image it can be seen that the given graph structure is similar to Rock-Paper-Scissor.
Input: arr[]={{“Snake”, “Water”}, {“Water”, “Gun”}, {“Gun”, “Gun”}}
Output: No
方法：根据观察有向图Rock-Paper-Scissor中只有三个节点，每个节点的入度和出度为一个，可以利用图的入度和出度来解决问题。请按照以下步骤解决问题。
- 初始化{ 字符串 , int} 的两个映射，比如inDegree和outDegree ，以存储这些映射中每个节点的入度和出度。
- 初始化一组字符串say st并插入所有节点，即该集合中的字符串。
- 检查任何地图的大小是否不等于3或节点集st的大小不等于3，然后打印“否”。
- 遍历两个映射并检查任何条目映射的第二个值是否不是1 ，然后打印“ No ”并中断。
- 最后，如果以上情况都不满足，则打印“是”。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if the given directed
// graph has same structure to Rock-Paper-Scissor
string similarGame(vector > arr)
{
// Stores indegree of each node
unordered_map indegree;
// Stores outdegree of each node
unordered_map outdegree;
// Stores all nodes
set st;
// Traverse through the edges
for (int i = 0; i < arr.size(); i++) {
// Check presence of self loop
if (arr[i][0] == arr[i][1])
return "No";
// Insert the node arr[i][0] and arr[i][1]
// into the set
st.insert(arr[i][0]);
st.insert(arr[i][1]);
// Increment the outdegree and indegree of
// nodes arr[i][0] and arr[i][1]
outdegree[arr[i][0]]++;
indegree[arr[i][1]]++;
}
// Check base conditions
if (outdegree.size() != 3 || indegree.size() != 3
|| st.size() != 3)
return "No";
// Traverse the array outdegree
for (auto it : outdegree) {
if (it.second != 1)
return "No";
}
// Traverse the array indegree
for (auto it : indegree) {
if (it.second != 1)
return "No";
}
// Return
return "Yes";
}
// Driver Code
int main()
{
// Given Input
vector > arr = { { "Snake", "Water" },
{ "Water", "Gun" },
{ "Gun", "Snake" } };
// Function Call
cout << similarGame(arr);
return 0;
} Python3
# Python 3 program for the above approach
# Function to check if the given directed
# graph has same structure to Rock-Paper-Scissor
def similarGame(arr):
# Stores indegree of each node
indegree = {}
# Stores outdegree of each node
outdegree = {}
# Stores all nodes
st = set()
# Traverse through the edges
for i in range(len(arr)):
# Check presence of self loop
if (arr[i][0] == arr[i][1]):
return "No"
# Insert the node arr[i][0] and arr[i][1]
# into the set
st.add(arr[i][0])
st.add(arr[i][1])
# Increment the outdegree and indegree of
# nodes arr[i][0] and arr[i][1]
if arr[i][0] in outdegree:
outdegree[arr[i][0]] += 1
else:
outdegree[arr[i][0]] = 0
if arr[i][1] in indegree:
indegree[arr[i][1]] += 1
else:
indegree[arr[i][1]] = 0
# Check base conditions
if (len(outdegree) != 3 and len(indegree) != 3 and len(st) != 3):
return "No";
# Traverse the array outdegree
for key,value in outdegree.items():
if (value == 1):
return "No"
# Traverse the array indegree
for key,value in indegree.items():
if (value == 1):
return "No"
# Return
return "Yes"
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [["Snake", "Water"],["Water", "Gun"],["Gun", "Snake"]]
# Function Call
print(similarGame(arr))
# This code is contributed by SURENDRA_GANGWAR.Yes时间复杂度： O(N)
辅助空间： O(N)
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