PAT 2018 秋

Posted 2022-08-02 codewars

A 1148 Werewolf - Simple Version

　　思路比较直接：模拟就行。因为需要序列号最小的两个狼人，所以以狼人为因变量进行模拟。

 

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 
 7 using namespace std;
 8 int N;
 9 vector<int> stateVec;
10 vector<int> inputVec;
11 bool judgeRight(int a, int b)
12 
13     int liarCnt = 0, wereLiarCnt = 0, tmpNum;
14     fill(stateVec.begin(), stateVec.end(), 0);
15     stateVec[a] = stateVec[b] = 1;
16     for(int i = 1; i <= N; ++ i)
17     
18         tmpNum = inputVec[i];
19         if((tmpNum > 0 && stateVec[tmpNum] == 1) || (tmpNum < 0 && stateVec[-tmpNum] == 0))
20         
21             liarCnt ++;
22             if(i == a || i == b)
23                 wereLiarCnt ++;
24         
25     
26     if(liarCnt == 2 && wereLiarCnt == 1)
27         return true;
28     else
29         return false;
30 
31 int main()
32 
33     cin >> N;
34     inputVec.resize(N+1, 0);
35     stateVec.resize(N+1, 0);
36     for(int i = 1; i <= N; ++ i)
37         cin >> inputVec[i];
38     for(int i = 1; i < N; ++i)
39     
40         for(int j = i+1; j <= N; ++ j)
41         
42             if(judgeRight(i,j))
43             
44                 cout << i << " " << j;
45                 return 0;
46             
47         
48     
49     cout << "No Solution";
50     return 0;
51 

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A 1149 Dangerous Goods Packaging

　　这个最开始使用图+二重for循环直接判，发现超时很严重。后来改用邻接表存储，每个物品更新一下与其互斥物品的标志。成功通过。

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <unordered_map>
 7 using namespace std;
 8 int N, M;
 9 typedef long long LL;
10 unordered_map<int, vector<int>> goodsMap;
11 vector<int> goodsVec;
12 bool judgeRight(void)
13 
14     int tmpNum;
15     int len = goodsVec.size();
16     unordered_map<int, int> incomFlagMap; 
17     for(int i = 0; i < len; ++ i)
18     
19         tmpNum = goodsVec[i];
20         if(incomFlagMap[tmpNum] == 1)
21             return false;
22         for(auto it = goodsMap[tmpNum].begin(); it != goodsMap[tmpNum].end(); ++it)
23             incomFlagMap[*it] = 1;
24     
25     return true;
26 
27 int main()
28 
29     cin >> N >> M;
30     int tmpNum2, tmpNum1;
31     while(N--)
32     
33         cin >> tmpNum1 >> tmpNum2;
34         goodsMap[tmpNum1].push_back(tmpNum2);
35         goodsMap[tmpNum2].push_back(tmpNum1);
36     
37     while(M--)
38     
39         cin >> N;
40         goodsVec.resize(N,0);
41         while(N--)
42             cin >> goodsVec[N];
43         if(judgeRight())
44             cout << "Yes" << endl;
45         else
46             cout << "No" << endl;
47     
48     return 0;
49 

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A 1150 Travelling Salesman Problem

　　这个理解题意就行。需要注意的地方：

　　　　　　　　　　　　　　1.不能通过的路径，距离输出为NA

　　　　　　　　　　　　　　2.最小距离是在TS cycle 和 TS simple cycle中选出

　　　　　　　　　　　　　　3.TS cycle 需要最后一个城市为出发城市

 

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 using namespace std;
 7 int N, M, K;
 8 const int INF = 0x7f7f7f7f;
 9 int route[210][210]=0;
10 int main()
11 
12     cin >> N >> M;
13     int tmpSt, tmpEnd, tmpDis, tmpNum, lastCity, nowCity, minDis = INF, minIndex;
14     while(M--)
15     
16         cin >> tmpSt >> tmpEnd >> tmpDis;
17         route[tmpSt][tmpEnd] = tmpDis;
18         route[tmpEnd][tmpSt] = tmpDis;
19     
20     cin >> K;
21     for(int i = 1; i <= K; ++i)
22     
23         tmpDis = 0;
24         cin >> tmpNum;
25         cin >> tmpSt;
26         lastCity = tmpSt;
27         vector<int> flagVec(N+1,0);
28         bool simpleFlag = true;
29         int travelCityCnt = 0;
30         for(int j = 1; j < tmpNum; j++)
31         
32             cin >> nowCity;
33             if(flagVec[nowCity] == 0)
34             
35                 flagVec[nowCity] = 1;
36                 travelCityCnt ++;
37             
38             else
39                 simpleFlag = false;
40             if(tmpDis >= 0 && route[lastCity][nowCity] > 0)
41                 tmpDis += route[lastCity][nowCity];
42             else
43                 tmpDis = -1;
44             lastCity = nowCity;
45         
46         if(lastCity == tmpSt && simpleFlag && tmpDis >= 0 && travelCityCnt == N)
47             printf("Path %d: %d (TS simple cycle)\n", i, tmpDis);
48         else if(tmpDis >= 0 && lastCity == tmpSt && travelCityCnt >= N)
49             printf("Path %d: %d (TS cycle)\n", i, tmpDis);
50         else
51         
52             tmpDis >= 0 ? printf("Path %d: %d (Not a TS cycle)\n", i, tmpDis) : printf("Path %d: NA (Not a TS cycle)\n", i);
53             tmpDis = -1;
54         
55         if(tmpDis > 0 && tmpDis < minDis)
56         
57             minDis = tmpDis;
58             minIndex = i;
59         
60     
61     printf("Shortest Dist(%d) = %d", minIndex, minDis);
62     return 0;
63 

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A 1151 LCA in a Binary Tree

 

　　只得到了22分，剩下的8分找不到原因了。╮(╯▽╰)╭

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <unordered_map>
 7 using namespace std;
 8 int N, M, K;
 9 vector<int> preOrderVec, inOrderVec;
10 unordered_map<int, int> valToIndexMap;
11 unordered_map<int, int> indexToValMap;
12 void insertVal(int inL, int inR, int preL, int preR, int index)
13 
14     if(inL > inR || preL > preR)
15         return;
16     int tmpNum = preOrderVec[preL];
17     int tmpIndex = inL;
18     while(inOrderVec[tmpIndex] != tmpNum)    
19         tmpIndex++;
20     //treeNodeVec[index] = tmpNum;
21     valToIndexMap[tmpNum] = index;
22     indexToValMap[index] = tmpNum;
23     insertVal(inL, tmpIndex-1, preL+1, preL+tmpIndex-inL, index*2);
24     insertVal(tmpIndex+1, inR, preR-inR+tmpIndex+1,preR, index*2+1);
25     return;
26 
27 void getNodeInfo(int a, int b)
28 
29     int tmpIndex1 = valToIndexMap[a], tmpIndex2 = valToIndexMap[b];
30     if(tmpIndex1 == 0 && tmpIndex2 == 0)
31         printf("ERROR: %d and %d are not found.\n", a, b);
32     else if(tmpIndex1 == 0 || tmpIndex2 == 0)
33         tmpIndex1 == 0 ? printf("ERROR: %d is not found.\n", a) : printf("ERROR: %d is not found.\n", b);
34     else
35     
36         while(tmpIndex1 != tmpIndex2)
37             tmpIndex1 > tmpIndex2 ? tmpIndex1 /= 2 : tmpIndex2 /= 2;
38         int tmpNum = indexToValMap[tmpIndex1];
39         if(tmpNum == a || tmpNum == b)
40             tmpNum == a ? printf("%d is an ancestor of %d.\n", a, b) : printf("%d is an ancestor of %d.\n", b, a);
41         else
42             printf("LCA of %d and %d is %d.\n", a, b, tmpNum);
43     
44 
45 int main()
46 
47     cin >> M >> N;
48     int tmpNum1, tmpNum2;
49     preOrderVec.resize(N);
50     inOrderVec.resize(N);
51     for(int i = 0; i < N; ++i)
52         cin >> inOrderVec[i];
53     for(int i = 0; i < N; ++ i)
54         cin >> preOrderVec[i];
55     insertVal(0, N-1, 0, N-1, 1);
56     while(M--)
57     
58         cin >> tmpNum1 >> tmpNum2;
59         getNodeInfo(tmpNum1, tmpNum2);
60     
61     return 0;
62 

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