Educational Codeforces Round 49（A，B，C，D）

A Palindromic Twist

字符串模拟，暴力check下。

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <deque>
 5 #include <stack>
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <string>
10 #include <cstring>
11 #include <fstream>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15
16 #define eps 1e-8
17 #define PI acos(-1.0)
18 #define INF 0x3f3f3f3f
19 #define FAST_IO ios::sync_with_stdio(false)
20
21 typedef long long LL;
22
23 bool check(int a,int b){
24     for(int i=a-1;i<=a+1;i++){
25         if(i==a) continue;
26         for(int j=b-1;j<=b+1;j++){
27             if(j==b) continue;
28             if(i==j) return true;
29         }
30     }
31     return false;
32 }
33
34 int main(){
35     FAST_IO;
36     int t;
37     cin>>t;
38     while(t--){
39         string s;
40         int n,f=0;
41         cin>>n>>s;
42         for(int i=0,j=n-1;i<j;i++,j--){
43             if(s[i]!=s[j]){
44                 if(check(s[i],s[j])) f=f;
45                 else {f=1;break;}
46             }
47         }
48         if(f==1) cout<<"NO"<<endl;
49         else cout<<"YES"<<endl;
50     }
51     return 0;
52 }

B Numbers on the Chessboard

分情况大讨论下。

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <deque>
 5 #include <stack>
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <string>
10 #include <cstring>
11 #include <fstream>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15
16 #define eps 1e-8
17 #define PI acos(-1.0)
18 #define INF 0x3f3f3f3f
19 #define FAST_IO ios::sync_with_stdio(false)
20
21 typedef long long LL;
22
23 int main(){
24     FAST_IO;
25     LL n,q;
26     cin>>n>>q;
27     for(LL i=1;i<=q;i++){
28         LL x,y;
29         cin>>x>>y;
30         if((x+y)%2==0){
31             if(n%2==0){
32                 cout<<(x-1)*n/2+(y+1)/2<<endl;
33             }
34             else{
35                 if((x-1)%2==0){
36                     cout<<(x-1)/2*n+(y+1)/2<<endl;
37                 }
38                 else{
39                     cout<<x/2*n-(n/2-(y+1)/2)<<endl;
40                 }
41             }
42         }
43         else{
44             if(n%2==0){
45                 cout<<n*n/2+(x-1)*n/2+(y+1)/2<<endl;
46             }
47             else{
48                 if((x-1)%2==0){
49                     cout<<n*n/2+1+(x-1)/2*n+(y+1)/2<<endl;
50                 }
51                 else{
52                     cout<<n*n/2+1+x/2*n-(n/2+1-(y+1)/2)<<endl;
53                 }
54             }
55         }
56     }
57     return 0;
58 }

C Minimum Value Rectangle

D Mouse Hunt

tarjan强连通分量，先缩点，在缩完的每个点中拿出度为0的点的最小的值。

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <deque>
 5 #include <stack>
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <string>
10 #include <cstring>
11 #include <fstream>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15
16 #define eps 1e-8
17 #define PI acos(-1.0)
18 #define INF 0x3f3f3f3f
19 #define FAST_IO ios::sync_with_stdio(false)
20
21 typedef long long LL;
22 const int N=2e5+10;
23 int scc,idx;
24 int dfn[N],low[N],ins[N],col[N],val[N],c[N],out[N];
25 stack <int> sk;
26 vector <int> g[N],dag[N];
27
28 void tarjan(int u){
29     dfn[u]=low[u]=++idx;
30     sk.push(u);
31     ins[u]=1;
32     int v;
33     for(int i=0;i<g[u].size();i++){
34         v=g[u][i];
35         if(!dfn[v]) tarjan(v);
36         if(ins[v]) low[u]=min(low[u],low[v]);
37     }
38     if(dfn[u]==low[u]){
39         scc++;
40         int mn=INF;
41         do{
42             v=sk.top();
43             sk.pop();
44             ins[v]=0;
45             col[v]=scc;
46             mn=min(mn,c[v]);
47         }while(u!=v);
48         val[scc]=mn;
49     }
50 }
51
52 int main(){
53     int n;
54     scanf("%d",&n);
55     for(int i=1;i<=n;i++) scanf("%d",&c[i]);
56     for(int i=1;i<=n;i++){
57         int cur;
58         scanf("%d",&cur);
59         g[i].push_back(cur);
60     }
61     for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i);
62     for(int i=1;i<=n;i++){
63         int u=col[i];
64         for(int j=0;j<g[i].size();j++){
65             int v=col[g[i][j]];
66             if(u!=v){
67                 dag[u].push_back(v);
68                 out[u]++;
69             }
70         }
71     }
72     int ans=0;
73     for(int i=1;i<=scc;i++){
74         if(out[i]==0) ans+=val[i];
75     }
76     printf("%d\n",ans);
77     return 0;
78 }

原文地址：https://www.cnblogs.com/ehanla/p/9527419.html