PAT 2014 秋

Posted 2022-08-12 codewars

A 1084 Broken Keyboard

　　注意大小写即可。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <string>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 
10     string inStr, outStr;
11     cin >> inStr >> outStr;
12     int inIndex = 0, outIndex = 0;
13     bool charFlag[200]=false;
14     while(inIndex < inStr.size())
15     
16         if(inStr[inIndex] == outStr[outIndex])
17         
18             inIndex++;
19             outIndex < outStr.size()-1 ? outIndex++ : outStr[outIndex] = ‘ ‘;
20         
21         else
22         
23             int tmpNum = (int)inStr[inIndex++];
24             tmpNum = tmpNum >= ‘a‘ ? tmpNum-‘a‘+‘A‘ : tmpNum;
25             if(!charFlag[tmpNum])
26             
27                 charFlag[tmpNum] = 1;
28                 printf("%c", tmpNum);  
29             
30         
31     
32     return 0;
33 

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A 1085 Perfect Sequence

　　1.二分查找，不然会超时

　　2.注意数值的范围，有些地方需要使用long long

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <vector>
 5 
 6 using namespace std;
 7 typedef long long LL;
 8 vector<int> inputNumVec;
 9 int N, P;
10 int getMaxCnt(int lIndex)
11 
12     int rIndex = inputNumVec.size()-1, midIndex;
13     LL tmpNum = (LL)inputNumVec[lIndex]*P;
14     while(lIndex <= rIndex)
15     
16         midIndex = (lIndex + rIndex) >> 1;
17         if(inputNumVec[midIndex] > tmpNum)
18             rIndex = midIndex - 1;
19         else
20             lIndex = midIndex + 1;
21     
22     return rIndex;
23 
24 int main()
25 
26     cin >> N >> P;
27     inputNumVec.resize(N);
28     for(int i = 0; i < N; ++ i)
29         cin >> inputNumVec[i];
30     sort(inputNumVec.begin(), inputNumVec.end());
31     int maxCnt = 0, tmpNum;
32     for(int i = 0; i <= inputNumVec.size()-maxCnt; ++ i)
33     
34         tmpNum = getMaxCnt(i)-i+1;
35         if(tmpNum > maxCnt)
36             maxCnt = tmpNum;
37     
38     cout << maxCnt;
39     return 0;
40 

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A 1086 Tree Traversals Again

　　自己想个办法建树之后进行后续遍历即可

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <string>
 6 #include <stack>
 7 
 8 using namespace std;
 9 typedef long long LL;
10 #define NoAddr -1
11 typedef struct NODE
12 
13     int val, lChild, rChild;
14     NODE(int v):val(v),lChild(NoAddr),rChild(NoAddr)
15 node;
16 vector<node> nodeVec;
17 stack<int> valVec;
18 bool coutFlag = false;
19 void postOrder(int tmpRoot)
20 
21     if(tmpRoot >= nodeVec.size())
22         return;
23     if(nodeVec[tmpRoot].lChild > 0)
24         postOrder(nodeVec[tmpRoot].lChild);
25     if(nodeVec[tmpRoot].rChild > 0)
26         postOrder(nodeVec[tmpRoot].rChild);
27     coutFlag ?  printf(" ") :coutFlag = true;
28     cout << nodeVec[tmpRoot].val;
29 
30 int main()
31 
32     int N, tmpNum, lastIndex = 0;
33     string tmpStr;
34     bool pushFlag = true;
35     cin >> N;
36     while(N > 0)
37     
38         cin >> tmpStr;
39         if(tmpStr == "Push")
40         
41             cin >> tmpNum;
42             valVec.push(nodeVec.size());
43             nodeVec.push_back(NODE(tmpNum));
44             pushFlag ? nodeVec[lastIndex].lChild = nodeVec.size()-1 :
45                        nodeVec[lastIndex].rChild = nodeVec.size()-1 ;
46             lastIndex = nodeVec.size()-1;
47             pushFlag = true;
48         
49         else
50         
51             lastIndex = valVec.top();
52             valVec.pop();
53             pushFlag = false;
54             N--;
55         
56     
57     postOrder(0);
58     return 0;
59 

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A 1087 All Roads Lead to Rome

　　1.城市名string 和 序号index的转换

　　2.Dijkstra找最短距离

　　3.DFS根据题目中给出的优先级找到所需的条件

  1 #include <cstdio>
  2 #include <stdlib.h>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <string>
  8 #include <stack>
  9 #include <unordered_map>
 10 using namespace std;
 11 #define NoAddr -1
 12 #define MAX_CITY 210
 13 #define CLR(a,b) memset(a,b,sizeof(a))
 14 typedef long long LL;
 15 const int INF = 0x7f7f7f7f;
 16 int N, K;
 17 string citySt;
 18 int route[MAX_CITY][MAX_CITY]=0;
 19 vector<int> preVec[MAX_CITY];
 20 vector<int> tmpPath;
 21 vector<int> outPath;
 22 vector<int> disVec;
 23 vector<int> cityHappyValVec;
 24 int minCityCnt = MAX_CITY, maxHappy = -1, minPathCnt = 0, tmpMaxHappy = 0;
 25 unordered_map<string, int> cityToIndexMap;
 26 unordered_map<int, string> indexToCityMap;
 27 void Dijkstra(int u)
 28 
 29     vector<bool> visitFlagVec(N, false);
 30     disVec.resize(N, INF);
 31     disVec[u] = 0;
 32     for(int i = 0; i < N; ++ i)
 33     
 34         int minDis = INF, v = -1;
 35         for(int j = 0; j < N; ++ j)
 36             if(!visitFlagVec[j] && disVec[j] < minDis)
 37             
 38                 minDis = disVec[j];
 39                 v = j;
 40             
 41         if(v == -1)
 42             return;
 43         visitFlagVec[v] = true;
 44         for(int j = 0; j < N; ++ j)
 45         
 46             if(!visitFlagVec[j] && disVec[v] + route[v][j] < disVec[j])
 47             
 48                 disVec[j] = disVec[v] + route[v][j];
 49                 preVec[j].clear();
 50                 preVec[j].push_back(v);
 51             
 52             else if(!visitFlagVec[j] && disVec[v] + route[v][j] == disVec[j])
 53             
 54                 preVec[j].push_back(v);
 55             
 56         
 57     
 58 
 59 void dfs(int u)
 60 
 61     if(u == 0)
 62     
 63         minPathCnt ++;
 64         tmpPath.push_back(0);
 65         if(tmpMaxHappy > maxHappy)
 66         
 67             maxHappy = tmpMaxHappy;
 68             outPath = tmpPath;
 69             minCityCnt = tmpPath.size();
 70         
 71         else if(tmpMaxHappy == maxHappy && tmpPath.size() < minCityCnt)
 72         
 73             outPath = tmpPath;
 74         
 75         tmpPath.pop_back();
 76     
 77     tmpPath.push_back(u);
 78     tmpMaxHappy += cityHappyValVec[u];
 79     for(int i = 0; i < preVec[u].size(); ++ i)
 80         dfs(preVec[u][i]);
 81     tmpMaxHappy -= cityHappyValVec[u];
 82     tmpPath.pop_back();
 83 
 84 int main()
 85 
 86     cin >> N >> K >> citySt;
 87     cityHappyValVec.resize(N, 0);
 88     string tmpCity, tmpSt, tmpEnd;
 89     int tmpHappy, tmpDis, stIndex, endIndex, endCityIndex;
 90     cityToIndexMap[citySt] = 0;
 91     indexToCityMap[0] = citySt;
 92     for(int i = 1; i < N; ++ i)
 93     
 94         cin >> tmpCity >> tmpHappy;
 95         cityHappyValVec[i] = tmpHappy;
 96         cityToIndexMap[tmpCity] = i;
 97         indexToCityMap[i] = tmpCity;
 98         if(tmpCity == "ROM")
 99             endCityIndex = i;
100     
101     CLR(route,0x7f);
102     while(K--)
103     
104         cin >> tmpSt >> tmpEnd >> tmpDis;
105         stIndex = cityToIndexMap[tmpSt];
106         endIndex = cityToIndexMap[tmpEnd];
107         route[stIndex][endIndex] = tmpDis;
108         route[endIndex][stIndex] = tmpDis;
109     
110     Dijkstra(0);
111     dfs(endCityIndex);
112     cout << minPathCnt << " " << disVec[endCityIndex] << " " << maxHappy << " " << (int)maxHappy/(outPath.size()-1) << endl;
113     bool symbolFlag = false;
114     for(int i = outPath.size()-1; i >= 0; -- i)
115     
116         symbolFlag ? printf("->") : symbolFlag = true;
117         cout << indexToCityMap[outPath[i]];
118     
119     return 0;
120 

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