【CodeForce】【#286】Div.2

T_T越来越水了，这次只做出A+B.

A题为了代码简单直接枚举（插入位置和插入字符）

 1 //CF #286 Div.2 A
 2 #include<vector>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<iostream>
 8 #include<algorithm>
 9 #define rep(i,n) for(int i=0;i<n;++i)
10 #define F(i,j,n) for(int i=j;i<=n;++i)
11 #define D(i,j,n) for(int i=j;i>=n;--i)
12 using namespace std;
13 const int N=10086;
14
15 void read(int &v){
16     v=0;int sign=1; char ch=getchar();
17     while(ch<‘0‘ || ch>‘9‘) {if (ch==‘-‘) sign=-1; ch=getchar();}
18     while(ch>=‘0‘&&ch<=‘9‘){v=v*10+ch-‘0‘; ch=getchar();}
19     v*=sign;
20 }
21 /****************tamplate***********************/
22
23 string s,s1;
24 bool check(int pos,char ch){
25     s1="";
26     rep(i,pos) s1+=s[i];
27     s1+=ch;
28     F(i,pos,s.length()-1) s1+=s[i];
29
30     for(int l=0,r=s1.length()-1;l<r;l++,r--)
31         if (s1[l]!=s1[r]) return 0;
32     return 1;
33 }
34
35 int main(){
36     ios::sync_with_stdio(false);
37     cin >> s;
38     int l=0,r=s.length(),pos=-1;
39
40     bool sign=1;
41     F(i,0,s.length())
42         rep(j,26) if (check(i,j+‘a‘)) {
43             cout <<s1<<endl;
44             return 0;
45         }
46     printf("NA");
47     return 0;
48 }

B题2B了一次，反正范围小，每种颜色的出边搜一遍就行了。

 1 //CF #286 Div.2 B
 2 #include<vector>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cstdlib>
 6 #include<vector>
 7 #include<iostream>
 8 #include<algorithm>
 9 #define rep(i,n) for(int i=0;i<n;++i)
10 #define F(i,j,n) for(int i=j;i<=n;++i)
11 #define D(i,j,n) for(int i=j;i>=n;--i)
12 #define pb push_back
13 using namespace std;
14 const int N=110;
15
16 void read(int &v){
17     v=0;int sign=1; char ch=getchar();
18     while(ch<‘0‘ || ch>‘9‘) {if (ch==‘-‘) sign=-1; ch=getchar();}
19     while(ch>=‘0‘&&ch<=‘9‘){v=v*10+ch-‘0‘; ch=getchar();}
20     v*=sign;
21 }
22 /****************tamplate***********************/
23 int n,m,cnt=0;
24 struct edge{
25     int from,to,color;
26 };
27 vector<edge>E;
28 vector<int>G[N];
29 void add(int x,int y,int z){
30     E.pb((edge){x,y,z});
31     E.pb((edge){y,x,z});
32     int m=E.size();
33     G[x].pb(m-2);
34     G[y].pb(m-1);
35 }
36
37 bool vis[N],yes[N];
38 void dfs(int x,int fa,int y,int col){
39     if (x==y) {yes[col]=1; return;}
40     vis[x]=1;
41     rep(i,G[x].size()){
42         edge&e=E[G[x][i]];
43         if (e.color==col && vis[e.to]==0) dfs(e.to,x,y,col);
44     }
45 }
46
47 int main(){
48     read(n); read(m);
49     int x,y,z;
50     F(i,1,m){
51         read(x); read(y); read(z);
52         add(x,y,z);
53     }
54     int Q;
55     read(Q);
56     F(i,1,Q){
57         read(x);
58         read(y);
59         memset(yes,0,sizeof yes);
60         rep(i,G[x].size()){
61             memset(vis,0,sizeof vis);
62             dfs(E[G[x][i]].to,x,y,E[G[x][i]].color);
63         }
64         int cnt=0;
65         F(i,1,m) cnt+=yes[i];
66         printf("%d\n",cnt);
67     }
68     return 0;
69 }