九度oj 题目1481：Is It A Tree?
Posted Jason杰
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- 题目描述：
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A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.
- 输入：
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The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero and less than 10000.
- 输出：
-
For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
- 样例输入：
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6 8 5 3 5 2 6 4 5 6 0 0 8 1 7 3 6 2 8 9 7 5 7 4 7 8 7 6 0 0 3 8 6 8 6 4 5 3 5 6 5 2 0 0 -1 -1
- 样例输出：
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Case 1 is a tree. Case 2 is a tree. Case 3 is not a tree.
一开始的代码如下1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 using namespace std; 7 8 int par[10002]; 9 int height[10002]; 10 11 void init() { 12 for (int i = 0; i < 10002; i++) { 13 par[i] = i; 14 height[i] = 0; 15 } 16 } 17 18 int find(int t) { 19 if (par[t] == t) { 20 return t; 21 } 22 return par[t] = find(par[t]); 23 } 24 25 void unite(int x, int y) { 26 x = find(x); 27 y = find(y); 28 if (x == y) { 29 return; 30 } 31 else { 32 if (height[x] < height[y]) { 33 par[x] = y; 34 } 35 else { 36 par[y] = x; 37 if (height[x] == height[y]) { 38 height[x]++; 39 } 40 } 41 } 42 } 43 44 bool isA(int p) { 45 int root = find(0); 46 for (int i = 1; i < p; i++) { 47 if (find(i) != root) { 48 return false; 49 } 50 } 51 return true; 52 } 53 54 map<int, int> mapping; 55 56 int main() { 57 //freopen("input.txt", "r", stdin); 58 //freopen("output.txt", "w", stdout); 59 int a, b; 60 int caseCnt = 0; 61 while (scanf("%d %d", &a,&b) != EOF && !(a == -1 && b ==-1)) { 62 int p = 0; 63 int cnt = 1; 64 init(); 65 mapping.clear(); 66 caseCnt++; 67 if (mapping.find(a) == mapping.end()) { 68 mapping[a] = p++; 69 } 70 if (mapping.find(b) == mapping.end()) { 71 mapping[b] = p++; 72 } 73 unite(mapping[a], mapping[b]); 74 while (scanf("%d %d", &a, &b) && !(a == 0 && b == 0)) { 75 if (mapping.find(a) == mapping.end()) { 76 mapping[a] = p++; 77 } 78 if (mapping.find(b) == mapping.end()) { 79 mapping[b] = p++; 80 } 81 unite(mapping[a], mapping[b]); 82 cnt++; 83 } 84 bool flag = isA(p); 85 flag = flag && (cnt == p - 1); 86 if (flag) { 87 printf("Case %d is a tree.\n", caseCnt); 88 } 89 else { 90 printf("Case %d is not a tree.\n", caseCnt); 91 } 92 93 } 94 return 0; 95 }
一开始思路比较简单，考虑如果图是连通的，那么如果边数为端点数减1，那么就是树
提交错误
开始没有考虑到可能有重边的情况，比如有两条a到b的边，或存在a到a的边，因此加一个判断，代码如下
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 using namespace std; 7 8 int par[10002]; 9 int height[10002]; 10 11 void init() { 12 for (int i = 0; i < 10002; i++) { 13 par[i] = i; 14 height[i] = 0; 15 } 16 } 17 18 int find(int t) { 19 if (par[t] == t) { 20 return t; 21 } 22 return par[t] = find(par[t]); 23 } 24 25 void unite(int x, int y) { 26 x = find(x); 27 y = find(y); 28 if (x == y) { 29 return; 30 } 31 else { 32 if (height[x] < height[y]) { 33 par[x] = y; 34 } 35 else { 36 par[y] = x; 37 if (height[x] == height[y]) { 38 height[x]++; 39 } 40 } 41 } 42 } 43 44 bool isSame(int x, int y) { 45 x = find(x); 46 y = find(y); 47 if (x == y) { 48 return true; 49 } 50 else { 51 return false; 52 } 53 } 54 55 bool isA(int p) { 56 int root = find(0); 57 for (int i = 1; i < p; i++) { 58 if (find(i) != root) { 59 return false; 60 } 61 } 62 return true; 63 } 64 65 map<int, int> mapping; 66 67 int main() { 68 //freopen("input.txt", "r", stdin); 69 //freopen("output.txt", "w", stdout); 70 int a, b; 71 int caseCnt = 0; 72 while (scanf("%d %d", &a,&b) != EOF && !(a == -1 && b ==-1)) { 73 int p = 0; 74 int cnt = 1; 75 init(); 76 mapping.clear(); 77 caseCnt++; 78 if (mapping.find(a) == mapping.end()) { 79 mapping[a] = p++; 80 } 81 if (mapping.find(b) == mapping.end()) { 82 mapping[b] = p++; 83 } 84 unite(mapping[a], mapping[b]); 85 bool flag = true; 86 while (scanf("%d %d", &a, &b) && !(a == 0 && b == 0)) { 87 if (mapping.find(a) == mapping.end()) { 88 mapping[a] = p++; 89 } 90 if (mapping.find(b) == mapping.end()) { 91 mapping[b] = p++; 92 } 93 if (isSame(a, b)) { 94 flag = false; 95 break; 96 } 97 unite(mapping[a], mapping[b]); 98 cnt++; 99 } 100 flag = flag && isA(p); 101 flag = flag && (cnt == p - 1); 102 if (flag) { 103 printf("Case %d is a tree.\n", caseCnt); 104 } 105 else { 106 printf("Case %d is not a tree.\n", caseCnt); 107 } 108 109 } 110 return 0; 111 }
在93行增加了一段代码，用于判断这种情况，提交依然错误。
又注意到题目中的边是有方向的，
比如5->6, 7->6 我的代码就判断不了
而且95行break掉会少读入数据
于是立马修改
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 using namespace std; 7 8 int par[10002]; 9 int height[10002]; 10 int cntD[10002]; 11 12 void init() { 13 for (int i = 0; i < 10002; i++) { 14 par[i] = i; 15 height[i] = 0; 16 } 17 } 18 19 int find(int t) { 20 if (par[t] == t) { 21 return t; 22 } 23 return par[t] = find(par[t]); 24 } 25 26 void unite(int x, int y) { 27 x = find(x); 28 y = find(y); 29 if (x == y) { 30 return; 31 } 32 else { 33 if (height[x] < height[y]) { 34 par[x] = y; 35 } 36 else { 37 par[y] = x; 38 if (height[x] == height[y]) { 39 height[x]++; 40 } 41 } 42 } 43 } 44 45 bool isSame(int x, int y) { 46 x = find(x); 47 y = find(y); 48 if (x == y) { 49 return true; 50 } 51 else { 52 return false; 53 } 54 } 55 56 bool isA(int p) { 57 int root = find(0); 58 for (int i = 1; i < p; i++) { 59 if (find(i) != root) { 60 return false; 61 } 62 } 63 return true; 64 } 65 66 map<int, int> mapping; 67 68 int main() { 69 //freopen("input.txt", "r", stdin); 70 //freopen("output.txt", "w", stdout); 71 int a, b; 72 int caseCnt = 0; 73 while (scanf("%d %d", &a,&b) != EOF && !(a == -1 && b ==-1)) { 74 int p = 0; 75 int cnt = 1; 76 init(); 77 mapping.clear(); 78 memset((void*)cntD, 0, sizeof(cntD)); 79 caseCnt++; 80 if (mapping.find(a) == mapping.end()) { 81 mapping[a] = p++; 82 } 83 if (mapping.find(b) == mapping.end()) { 84 mapping[b] = p++; 85 } 86 cntD[mapping[b]]++; 87 88 unite(mapping[a], mapping[b]); 89 bool flag = true; 90 while (scanf("%d %d", &a, &b) && !(a == 0 && b == 0)) { 91 if (flag == false) { 92 continue; 93 } 94 if (mapping.find(a) == mapping.end()) { 95 mapping[a] = p++; 96 } 97 if (mapping.find(b) == mapping.end()) { 98 mapping[b] = p++; 99 } 100 if (cntD[mapping[b]] == 1) { 101 flag = false; 102 continue; 103 } 104 if (isSame(a, b)) { 105 flag = false; 106 continue; 107 } 108 cntD[mapping[b]]++; 109 unite(mapping[a], mapping[b]); 110 cnt++; 111 } 112 flag = flag && isA(p); 113 flag = flag && (cnt == p - 1); 114 if (flag) { 115 printf("Case %d is a tree.\n", caseCnt); 116 } 117 else { 118 printf("Case %d is not a tree.\n", caseCnt); 119 } 120 121 } 122 return 0; 123 }
增加了一个记录每个端点入度的数组，如果其值大于1，那么就不是一颗树。
提交依然错误。。。。
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浏览了一下题解，终于发现问题
注意到题目第一行
A tree is a well-known data structure that is either empty (null, void, nothing）
树可以是空的，
也就是说如果他直接输入0 0，你要告诉它这是一棵树
再次修改代码
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 using namespace std; 7 8 int par[10002]; 9 int height[10002]; 10 int cntD[10002]; 11 12 void init() { 13 for (int i = 0; i < 10002; i++) { 14 par[i] = i; 15 height[i] = 0; 16 } 17 } 18 19 int find(int t) { 20 if (par[t] == t) { 21 return t; 22 } 23 return par[t] = find(par[t]); 24 } 25 26 void unite(int x, int y) { 27 x = find(x); 28 y = find(y); 29 if (x == y) { 30 return; 31 } 32 else { 33 if (height[x] < height[y]) { 34 par[x] = y; 35 } 36 else { 37 par[y] = x; 38 if (height[x] == height[y]) { 39 height[x]++; 40 } 41 } 42 } 43 } 44 45 bool isSame(int x, int y) { 46 x = find(x); 47 y = find(y); 48 if (x == y) { 49 return true; 50 } 51 else { 52 return false; 53 } 54 } 55 56 bool isA(int p) { 57 int root = find(0); 58 for (int i = 1; i < p; i++) { 59 if (find(i) != root) { 60 return false; 61 } 62 } 63 return true; 64 } 65 66 map<int, int> mapping; 67 68 int main() { 69 //freopen("input.txt", "r", stdin); 70 //freopen("output.txt", "w", stdout); 71 int a, b; 72 int caseCnt = 0; 73 while (scanf("%d %d", &a,&b) != EOF && !(a == -1 && b ==-1)) { 74 int p = 0; 75 int cnt = 1; 76 init(); 77 mapping.clear(); 78 memset((void*)cntD, 0, sizeof(cntD)); 79 caseCnt++; 80 81 if (a == 0 && b == 0) { 82 printf("Case %d is a tree.\n", caseCnt); 83 continue; 84 } 85 if (mapping.find(a) == mapping.end()) { 86 mapping[a] = p++; 87 } 88 if (mapping.find(b) == mapping.end()) { 89 mapping[b] = p++; 90 } 91 cntD[mapping[b]]++; 92 93 unite(mapping[a], mapping[b]); 94 bool flag = true; 95 while (scanf("%d %d", &a, &b) && !(a == 0 && b == 0)) { 96 if (flag == false) { 97 continue; 98 } 99 if (mapping.find(a) == mapping.end()) { 100 mapping[a] = p++; 101 } 102 if (mapping.find(b) == mapping.end()) { 103 mapping[b] = p++; 104 } 105 if (cntD[mapping[b]] == 1) { 106 flag = false; 107 continue; 108 } 109 if (isSame(a, b)) { 110 flag = false; 111 continue; 112 } 113 cntD[mapping[b]]++; 114 unite(mapping[a], mapping[b]); 115 cnt++; 116 } 117 flag = flag && isA(p); 118 flag = flag && (cnt == p - 1); 119 if (flag) { 120 printf("Case %d is a tree.\n", caseCnt); 121 } 122 else { 123 printf("Case %d is not a tree.\n", caseCnt); 124 } 125 126 } 127 return 0; 128 }
提交终于通过了。
还是要仔细读题呀！！！！！
后来又想到是不是直接判断入度即可判断出它是不是一棵树，写出了如下代码
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 using namespace std; 7 8 int cntD[10002]; 9 10 map<int, int> mapping; 11 12 bool isTree(int p) { 13 int t = 0; 14 for(int i = 0; i < p; i++) { 15 if(cntD[i] == 0) { 16 t++; 17 if(t > 1) { 18 return false; 19 } 20 } 21 else if(cntD[i] != 1) { 22 return false; 23 } 24 } 25 if(t == 0) { 26 return false; 27 } 28 return true; 29 } 30 int main() { 31 //freopen("input.txt", "r", stdin); 32 //freopen("output.txt", "w", stdout); 33 int a, b; 34 int caseCnt = 0; 35 while (scanf("%d %d", &a,&b) != EOF && !(a == -1 && b ==-1)) { 36 int p = 0; 37 int cnt = 1; 38 mapping.clear(); 39 memset((void*)cntD, 0, sizeof(cntD)); 40 caseCnt++; 41 42 if (a == 0 && b == 0) { 43 printf("Case %d is a tree.\n", caseCnt); 44 continue; 45 } 46 if (mapping.find(a) == mapping.end()) { 47 mapping[a] = p++; 48 } 49 if (mapping.find(b) == mapping.end()) { 50 mapping[b] = p++; 51 } 52 cntD[mapping[b]]++; 53 54 bool flag = true; 55 while (scanf("%d %d", &a, &b) && !(a == 0 && b == 0)) { 56 if (flag == false) { 57 continue; 58 } 59 if (mapping.find(a) == mapping.end()) { 60 mapping[a] = p++; 61 } 62 if (mapping.find(b) == mapping.end()) { 63 mapping[b] = p++; 64 } 65 if (cntD[mapping[b]] == 1) { 66 flag = false; 67 continue; 68 } 69 cntD[mapping[b]]++; 70 cnt++; 71 } 72 flag = flag && isTree(p); 73 if (flag) { 74 printf("Case %d is a tree.\n", caseCnt); 75 } 76 else { 77 printf("Case %d is not a tree.\n", caseCnt); 78 } 79 80 } 81 return 0; 82 }
提交答案错误。
这里存在一种特殊情况
比如输入 2 1 3 4 4 5 5 3 0 0
即 2->1 3->4->5->3
成环时也满足只有一个入度为0，其余入度均为1
因此还需要去判断是否成环，有点麻烦，不想做了
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