CodeForces - 919D Substring （DP 记忆化搜索）

最新推荐文章于 2020-11-11 00:16:56 发布

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最新推荐文章于 2020-11-11 00:16:56 发布

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Why is everything so heavy?

本文链接：https://blog.youkuaiyun.com/lzc504603913/article/details/79223514

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本文介绍了一种通过记忆化搜索解决图中寻找具有最大字母出现次数路径的问题，并提供了完整的C++实现代码。该算法适用于有向图，通过深度优先搜索（DFS）结合动态规划来求解每个节点所能达到的最大相同字母的频率。

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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 4
abaca
1 2
1 3
3 4
4 5

output
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.

解题思路：记忆化搜索即可。dp[i][j]代表从第i个节点出发，每一位最多有多少个。用深搜即可。拓扑排序判环。。

#include <iostream>
#include <string>
#include <algorithm>
#include <map>
#include <memory.h>
#include <stack>

using namespace std;

inline void scan_d(int &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

struct edge
{
    int v1;
    int v2;
    int next;
}e[300005];
int head[300005];
int edge_num=0;


void insert(int v1,int v2)
{
    e[edge_num].v1=v1;
    e[edge_num].v2=v2;
    e[edge_num].next=head[v1];
    head[v1]=edge_num++;
}
char str[300005];


int dp[300005][30];
bool vis[300005];


void dfs(int x){

    vis[x]=1;
    for(int i=head[x];i!=-1;i=e[i].next){
        if(vis[e[i].v2]){
           // cout<<"a:  ";
            for(int j=0;j<26;j++){
                dp[x][j]=max(dp[e[i].v2][j],dp[x][j]);
              //  cout<<dp[x][j]<<" ";
            }
          //  cout<<endl;
        }
        else{
            dfs(e[i].v2);
           // cout<<"b:  ";
            for(int j=0;j<26;j++){
                dp[x][j]=max(dp[e[i].v2][j],dp[x][j]);
             //   cout<<dp[x][j]<<" ";
            }
           // cout<<endl;
        }
    }
    dp[x][str[x]-'a']++;
}

int indegree[300005];

bool isLoop(int N)   //G是邻接表
{
    stack<int> s;
    int i,k;
    for(i=1;i<=N;i++)
    {
        if(!indegree[i])
            s.push(i);     //入度为零的节点入栈
    }
    int count=0;

    while(!s.empty())
    {
        i = s.top();
        s.pop();
        count++;  //统计遍历过的节点个数
        for(int p=head[i];~p;p=e[p].next)
        {

            indegree[e[p].v2]--;
            if(!indegree[e[p].v2])
                s.push(e[p].v2);
        }
    }
    if(count==N)
        return false;
    return true;
}

int main()
{
    int n,m;
    scan_d(n);
    scan_d(m);
    //scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    int a,b;
    edge_num=0;
    memset(head,-1,sizeof(head));
    for(int i=0;i<m;i++){
        scan_d(a);
        scan_d(b);
      //  scanf("%d%d",&a,&b);
        insert(a,b);
        indegree[b]++;
    }

    if(isLoop(n)){
        cout<<-1<<endl;
    }
    else{

        for(int i=1;i<=n;i++)
            if(vis[i]==0)
                dfs(i);

        int ans=0;
        for(int i=1;i<=n;i++)
            for(int j=0;j<26;j++)
                ans=max(ans,dp[i][j]);

        cout<<ans<<endl;
    }

    return 0;

}