InputThe first line contains only one integer T (T≤100T≤100), which indicates the number of test cases.
For each test case, the first line contains only one integer n (2≤n≤1092≤n≤109), indicating the number of vertices.
OutputFor each test case, output one line "Case #x：y",where x is the case number (starting from 1) and y is the total weight of the minimum spanning tree.
Sample Input
2 2 3
Sample Output
Case #1: 2 Case #2: 5
代码：
#include <bits/stdc++.h> #define ll long long using namespace std; int main() { int T,tot=0; cin>>T; while(T--) { ll n; scanf("%lld",&n); ll ans=(n+1)*n/2; ans--; printf("Case #%d: %lld\n",++tot,ans); } return 0; }
Inputfirst line contains only one integer T (T≤125T≤125), which indicates the number of test cases. Each test case contains two integers A and B (1≤A≤B≤10181≤A≤B≤1018).OutputFor each test case, first output one line "Case #x:", where x is the case number (starting from 1).
Then in a new line, print an integer s indicating the number of pairs you find.
In each of the following s lines, print a pair of integers C and D. pairs should be sorted by C, and then by D in ascending order.
Sample Input
2 10 10 9 27
Sample Output
Case #1: 1 10 10 Case #2: 2 9 27 27 9
代码：
#include <bits/stdc++.h> using namespace std; int T; long long A, B, C, D; int main() { scanf("%d", &T); for(int t = 1; t <= T; t ++) { scanf("%lld%lld", &A, &B); printf("Case #%d:\n", t); if(A == B) { printf("1\n"); printf("%lld %lld\n", A, B); } else { printf("2\n"); printf("%lld %lld\n", A, B); printf("%lld %lld\n", B ,A); } }// return 0; }
InputThe first line contains only one integer T (T≤500T≤500), which indicates the number of test cases.
For each test case, the first line contains two integers n and m (1≤n,m≤301≤n,m≤30).
In the next n lines, each line contains m integers, j-th number in the i-th line means the symbol on the grid(the same number means the same symbol on the grid).
OutputFor each test case, there should be one line in the output.
You should output “Case #x: y”，where x is the case number(starting from 1), and y is a string representing the answer of the question. If there are at least one pair of grids that can be removed together, the y is “Yes”（without quote), else y is “No”.Sample Input
2 3 3 1 2 1 2 1 2 1 2 1 3 3 1 2 3 2 1 2 3 2 1
Sample Output
Case #1: Yes Case #2: No
代码：
#include <bits/stdc++.h> using namespace std; int a[35][35]; map <int,int> Map; int main() { int T,tot=0; cin>>T; while(T--) { Map.clear(); int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { scanf("%d",&a[i][j]); } } int flag=0; for(int i=1;i<=m;i++) { if(!Map[a[1][i]]) Map[a[1][i]]=1; else flag++; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } Map.clear(); for(int i=1;i<=m;i++) { if(!Map[a[n][i]]) Map[a[n][i]]=1; else flag++; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } Map.clear(); for(int i=1;i<=n;i++) { if(!Map[a[i][1]]) Map[a[i][1]]=1; else flag++; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } Map.clear(); for(int i=1;i<=n;i++) { if(!Map[a[i][m]]) Map[a[i][m]]=1; else flag++; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } for(int i=1;i<=n;i++) { for(int j=1;j<=m-1;j++) { if(a[i][j]==a[i][j+1]) { flag++; } } if(flag) break; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } for(int i=1;i<=m;i++) { for(int j=1;j<=n-1;j++) { if(a[j][i]==a[j+1][i]) { flag++; } } if(flag) break; } if(flag) { printf("Case #%d: Yes\n",++tot); continue; } else { printf("Case #%d: No\n",++tot); } } return 0; }
InputThe first line contains only one integer T (T≤1000T≤1000), which indicates the number of test cases.
For each test case, the first line contains two integers n (1≤n≤1000001≤n≤100000), q (0≤q≤1000000≤q≤100000).
In the following n -1 lines, the i-th line contains two integers ui,vi(1≤ui,vi≤n)ui,vi(1≤ui,vi≤n)indicating there is an edge between uiuii and vivi in the tree.
In the next q lines, the i-th line first comes with an integer mi(1≤mi≤100000)mi(1≤mi≤100000)indicating the number of vertices in the query set.Then comes with mi different integers, indicating the nodes in the query set.
It is guaranteed that ∑qi=1mi≤100000∑i=1qmi≤100000.
It is also guaranteed that the number of test cases in which n≥1000n≥1000 or ∑qi=1mi≥1000∑i=1qmi≥1000 is no more than 10.
OutputFor each test case, first output one line "Case #x:", where x is the case number (starting from 1).
Then q lines follow, i-th line contains an integer indicating the size of the auxiliary set for each query.
Sample Input
1 6 3 6 4 2 5 5 4 1 5 5 3 3 1 2 3 1 5 3 3 1 4
Sample Output
Case #1: 3 6 3
代码：
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 10; int T; int a[maxn]; int deep[maxn], fa[maxn], son[maxn], soon[maxn]; vector<int> v[maxn]; void dfs(int child, int parent) { fa[child] = parent; deep[child] = deep[parent] + 1; son[child] = 0; for(int i = 0; i < v[child].size(); i ++) { if(v[child][i] != parent) { dfs(v[child][i], child); son[child] ++; } } } bool cmp(int a, int b) { return deep[a] > deep[b]; } int main() { scanf("%d", &T); for(int t = 1; t <= T; t ++) { int N, M; scanf("%d%d", &N, &M); for(int i = 1; i <= N; i ++) v[i].clear(); for(int i = 0; i < N - 1; i ++) { int st, en; scanf("%d%d", &st, &en); v[st].push_back(en); v[en].push_back(st); } printf("Case #%d:\n", t); dfs(1, 0); while(M --) { int K; scanf("%d", &K); int ans = N - K; for(int i = 0; i < K; i ++) { scanf("%d", &a[i]); } sort(a, a + K, cmp); for(int i = 0; i < K; i ++) soon[a[i]] = son[a[i]]; for(int i = 0; i < K; i ++) { if(soon[a[i]] >= 2) ans ++; else if(soon[a[i]] == 0) soon[fa[a[i]]] --; } printf("%d\n", ans); } } return 0; }
原文：https://www.cnblogs.com/zlrrrr/p/10603025.html
