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 }
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 }
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 }
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 }
原文地址：https://www.cnblogs.com/codewars/p/11391781.html
时间： 2024-09-29 13:31:18