题目链接：Codeforces 467D Fedor and Essay

题目大意：给定一个含n个单词的文本，然后给定m种变换，要求变换后r的个数尽量少，长度尽量短，不区分大小写。

解题思路：bfs，将每个单词处理成长度以及r的个数，然后从最优的开始更新即可，类似dp。

#include <cstdio>
#include <cstring>
#include <map>
#include <string>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>

using namespace std;
const int maxn = 1e5+5;

typedef long long ll;
typedef pair<ll, ll> pii;

int M, N, sz, W[maxn];
map<string, int> V;
vector<int> g[maxn * 3];
pii vec[maxn*3];

void add (string& s) {
    ll len = s.length(), cnt = 0;

    for (int j = 0; j < len; j++) {
        if (s[j] >= ‘A‘ && s[j] <= ‘Z‘)
            s[j] = s[j] - ‘A‘ + ‘a‘;

        if (s[j] == ‘r‘)
            cnt++;
    }

    if (!V.count(s)) {
        V[s] = sz;
        vec[sz++] = make_pair(cnt, len);
    }
}

void init () {
    sz = 0;

    string s, e;

    cin >> M;
    for (int i = 0; i < M; i++) {
        cin >> s;
        add(s);
        W[i] = V[s];
    }

    cin >> N;
    for (int i = 0; i <    N; i++) {
        cin >> s >> e;
        add(s);
        add(e);
        g[V[e]].push_back(V[s]);
    }
}

void solve () {
    queue<int> que;
    for (int i = 0; i < sz; i++)
        que.push(i);

    while (!que.empty()) {
        int idx = que.front();
        pii u = vec[idx];
        que.pop();

        for (int i = 0; i < g[idx].size(); i++) {
            int v = g[idx][i];
            if (vec[v] > u) {
                vec[v] = u;
                que.push(v);
            }
        }
    }

    ll len = 0, cnt = 0;
    for (int i = 0; i < M; i++) {
        cnt += vec[W[i]].first;
        len += vec[W[i]].second;
    }
    cout << cnt << " " << len << endl;
}

int main () {
    init();
    solve();
    return 0;
}

Codeforces 467D Fedor and Essay(bfs)