D. Substring
Input
The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.
The second line contains a string s with only lowercase English letters. The i-th character is the letter assigned to the i-th node.
Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between xand y. Also the graph can be not connected.
Output
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
Examples
input
5 4
abaca
1 2
1 3
3 4
4 5
3
input
6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4
output
-1
input
10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7
output
4
Note
In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter 'a' appears 3 times.
题意 ：给你一个n长度的字符串 和一个m条边的图 字符串是图的顶点上的字符 问在图上的一条路上有几个相同字符 输出最多的字符个数，有环输 -1
1 拓扑排序 +dfs 搜索
搜索每个顶点的路上出现每个字母出现的最大次数（次数可能为0 因为这TLE了好多发），输出最大的就行
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 int flag=0; 5 vector<int>q[300007]; 6 int vi[300007][29]; 7 char s[300007]; 8 int ru[300007]; 9 #define rd(a) scanf("%d",&a) 10 int dfs2(int i,char c) 11 { 12 if(vi[i][c-'a']!=-1) return vi[i][c-'a']; 13 int maxx=0; 14 for(int j=0;j<q[i].size();j++) 15 { 16 17 maxx=max(dfs2(q[i][j],c),maxx); 18 } 19 if(s[i]==c) return vi[i][c-'a']=maxx+1; 20 else return vi[i][c-'a']=maxx; 21 22 } 23 int main() 24 { 25 while(~scanf("%d%d",&n,&m)) 26 { 27 28 scanf(" %s",s+1); 29 for(int i=0;i<m;i++) 30 { 31 int u,v; 32 rd(u),rd(v); 33 ru[v]++; 34 q[u].push_back(v); 35 } 36 stack<int>qq; 37 int cnt=0; 38 for(int i=1;i<=n;i++) 39 { 40 41 if(!ru[i]) 42 { 43 qq.push(i);cnt++; 44 } 45 } 46 47 while(!qq.empty()) 48 { 49 int v=qq.top(); 50 qq.pop(); 51 for(int i=0;i<q[v].size();i++) 52 { 53 ru[q[v][i]]--; 54 if(!ru[q[v][i]]) qq.push(q[v][i]),cnt++;; 55 } 56 } 57 if(cnt<n) 58 { 59 printf("-1 "); return 0; 60 } 61 memset(vi,-1,sizeof(vi)); 62 int maxx=0; 63 for(int i=1;i<=n;i++) 64 { 65 for(int j='a';j<='z';j++) 66 { 67 maxx=max(maxx,dfs2(i,j)); 68 } 69 70 } 71 printf("%d ",maxx); 72 73 74 } 75 76 77 78 return 0; 79 } 80 81 82 83 84
2 dfs判环 +dfs 搜索
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 int vis[300007]; 5 int flag=0; 6 vector<int>q[300007]; 7 int vi[300007][29]; 8 char s[300007]; 9 bool dfs(int a) 10 { 11 if(vis[a]==2) return true; 12 if(vis[a]==1) return false; 13 vis[a]=1; 14 for(int i=0;i<q[a].size();i++) 15 { 16 if(!dfs(q[a][i])) return false; 17 } 18 vis[a]=2; 19 return true; 20 } 21 #define rd(a) scanf("%d",&a) 22 int dfs2(int i,char c) 23 { 24 if(vi[i][c-'a']!=-1) return vi[i][c-'a']; 25 int maxx=0; 26 for(int j=0;j<q[i].size();j++) 27 { 28 29 maxx=max(dfs2(q[i][j],c),maxx); 30 } 31 if(s[i]==c) return vi[i][c-'a']=maxx+1; 32 else return vi[i][c-'a']=maxx; 33 34 } 35 int main() 36 { 37 while(~scanf("%d%d",&n,&m)) 38 { 39 40 scanf(" %s",s+1); 41 for(int i=0;i<m;i++) 42 { 43 int u,v; 44 rd(u),rd(v); 45 46 q[u].push_back(v); 47 } 48 for(int i=1;i<=n;i++) 49 { 50 if(!vis[i]&&!dfs(i)) 51 { 52 printf("-1 "); 53 return 0; 54 } 55 56 57 } 58 memset(vi,-1,sizeof(vi)); 59 int maxx=0; 60 for(int i=1;i<=n;i++) 61 { 62 for(int j='a';j<='z';j++) 63 { 64 maxx=max(maxx,dfs2(i,j)); 65 } 66 67 } 68 printf("%d ",maxx); 69 70 71 } 72 73 74 75 return 0; 76 }