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 }
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 }
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 }
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 }
原文地址：https://www.cnblogs.com/codewars/p/11374301.html
时间： 2024-10-07 16:07:09