Codeforces 919 D. Substring （記憶化搜尋）__動態規劃

Description

You are given a graph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path’s value as the number of the most frequently occurring letter. For example, if letters on a path are “abaca”, then the value of that path is 3. Your task is find a path whose value is the largest.

  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 x and 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 4abaca1 21 33 44 5

  Examples output

3

  題意

在圖中找一條路徑，滿足該路徑上某字母的出現頻率最高，輸出該字母的出現次數。（圖中可能存在自環以及重邊）

  思路

記憶化搜尋， dp[o][ch] d p [ o ] [ c h ] dp[o][ch] 代表節點 o o o 及以下字元 ch c h ch 出現的最大次數。

tarjan 判環，注意存在自環情況。

  AC 代碼

#include <bits/stdc++.h>#define IO ios::sync_with_stdio(false);\    cin.tie(0);\    cout.tie(0);using namespace std;const int maxn = 3e5+10;const int mod = 1e9+7;typedef __int64 LL;#define inf 0x7f7f7fstruct node{    int to;    int next;} edge[maxn];int head[maxn],tot,dfn[maxn],low[maxn];int dp[maxn][26],idx;int ch[maxn],in[maxn];bool instack[maxn];int Stack[maxn],top;int n,m;bool flag;char str[maxn];void init(){    memset(dp,0,sizeof dp);    memset(dfn,0,sizeof dfn);    memset(low,0,sizeof low);    memset(instack,false,sizeof instack);    memset(head,-1,sizeof head);    top = idx = tot = 0;    flag = true;}void addedge(int u,int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void dfs(int x){    dfn[x] = low[x] = ++idx;    instack[x] = true;    Stack[top++] = x;    for(int i=head[x]; i!=-1; i=edge[i].next)    {        int to = edge[i].to;        if(!dfn[to])        {            dfs(to);            low[x] = min(low[x],low[to]);        }        else if(instack[to] && dfn[to]<low[x])            low[x] = dfn[to];        for(int j=0; j<26; j++)            dp[x][j] = max(dp[x][j],dp[to][j]);    }    if(dfn[x]==low[x])    {        int Size = 0,now;        do        {            now = Stack[--top];            instack[now] = false;            ++Size;        }        while(now!=x);        if(Size>1)            flag = false;    }    if(!flag)        return;    ++dp[x][ch[x]];}void solve(){    for(int i=1; i<=n; i++)        ch[i] = str[i-1] - 'a';    int ans = -1;    for(int i=1; i<=n; i++)        if(in[i]==0)        {            dfs(i);            for(int j=0; j<26; j++)                ans = max(ans,dp[i][j]);        }    cout<<(flag?ans:-1)<<endl;}int main(){    IO;    init();    cin>>n>>m;    cin>>str;    for(int i=0; i<m; i++)    {        int u,v;        cin>>u>>v;        if(u==v)            flag = false;        ++in[v];        addedge(u,v);    }    solve();    return 0;}