cf 919D Substring

一 原题

                                                             D. Substring

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 35 44 36 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.

二 分析

DFS判断是否有环，如果有环，那么输出-1。如果没有环，那么DFS后得到原图的一个拓扑排序，对于任意一条图上的路径（v1->v2->v3...），在拓扑排序中保证v_i在v_i+1之前，那么我们就可以动归得到每个点作为起点时，某个字母最多出现多少次。下面两份代码，一份是DFS，一份是裸的拓扑排序。复杂度：O(26*(n+m))

三 代码

/*
PROB: cf 919D
LANG: c++
AUTHOR: maxkibble
*/

#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

// #define local
#define pb push_back
#define mp make_pair
#define pii pair<int, int>

const int maxn = 3e5 + 10;

int n, m, s, e, ans = 1, dp[maxn];
char ch[maxn];
bool in[maxn], vis[maxn];
vector<int> edge[maxn], seq;

bool circle(int x) {
bool ret = false;
in[x] = vis[x] = true;
for(auto nxt: edge[x]) {
if(in[nxt]) {
ret = true;
break;
}
if(!vis[nxt] && circle(nxt)) {
ret = true;
break;
}
}
in[x] = false;
seq.pb(x);
return ret;
}

int main() {
#ifdef local
freopen("d.in", "r", stdin);
#endif
scanf("%d%d%s", &n, &m, ch + 1);
for(int i = 0; i < m; i++) {
scanf("%d%d", &s, &e);
edge[s].pb(e);
}
for(int i = 1; i <= n; i++) {
if(vis[i]) continue;
if(circle(i)) {
puts("-1");
return 0;
}
}
for(char c = 'a'; c <= 'z'; c++) {
for(int i = 0; i < seq.size(); i++) {
int v = seq[i];
int flag = (ch[v] == c);
dp[v] = flag;
for(auto nxt: edge[v]) {
dp[v] = max(dp[v], dp[nxt] + flag);
}
ans = max(ans, dp[v]);
}
}
printf("%d\n", ans);
return 0;
}

/*
PROB: cf 919D
LANG: c++
AUTHOR: maxkibble
*/

#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#i
abf1
nclude <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

// #define local
#define pb push_back
#define mp make_pair
#define pii pair<int, int>

const int maxn = 3e5 + 10;

int n, m, s, e, ans = 1, degree[maxn], dp[maxn];
char ch[maxn];
queue<int> q;
vector<int> edge[maxn], seq;

void topSort() {
for(int i = 1; i <= n; i++) {
for(auto nxt: edge[i])
degree[nxt]++;
}
for(int i = 1; i <= n; i++) {
if(degree[i] == 0)
q.push(i);
}
while(!q.empty()) {
int u = q.front();
seq.pb(u);
q.pop();
for(auto v: edge[u]) {
degree[v]--;
if(degree[v] == 0)
q.push(v);
}
}
}

int main() {
#ifdef local
freopen("d.in", "r", stdin);
#endif
scanf("%d%d%s", &n, &m, ch + 1);
for(int i = 0; i < m; i++) {
scanf("%d%d", &s, &e);
edge[s].pb(e);
}

topSort();

if(seq.size() != n) {
puts("-1");
return 0;
}

for(char c = 'a'; c <= 'z'; c++) {
for(int i = n - 1; i >= 0; i--) {
int v = seq[i];
int flag = (ch[v] == c);
dp[v] = flag;
for(auto nxt: edge[v]) {
dp[v] = max(dp[v], dp[nxt] + flag);
}
ans = max(ans, dp[v]);
}
}
printf("%d\n", ans);
return 0;
}