题目链接：

题意：

给定n个单词。

下面有m个替换方式，左边的单词能变成右边的单词。

替换任意次后使得最后字母r个数最少，在r最少的情况下单词总长度最短

输出字母r的个数和单词长度。

思路：

我们认为一个单词有2个参数，则m个替换规则可以当成m个点的有向图。

则某些单词的替换终点会确定，所以反向建图bfs一下。

为了防止某些点被反复更新，所以把每个点的权值都放到栈里排个序然后bfs。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<string>
using namespace std;

typedef int ll;
const int inf = 1000000;
#define N 300005
int n, m;
void go(string &x){
    int len = x.length();
    for(int i = 0; i < len; i++)
        if('A' <= x[i] && x[i] <= 'Z')
            x[i] = x[i]-'A'+'a';
}
string s[N], a[N], b[N];
int S[N], A[N], B[N], LEN[N], NUM[N];
map<string, int> mp;
struct Edge{
    int to, nex;
}edge[N*10];
int head[N], edgenum;
void init(){memset(head, -1, sizeof head); edgenum = 0;}
void add(int u, int v){
    Edge E = {v, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
struct node{
    int x, y, point;
    node (int X = 0, int Y = 0, int P = 0):x(X), y(Y), point(P){}
    bool operator<(const node&D){
        if(D.x!=x)return D.x > x;
        return D.y>y;
    }
}st[N], dis[N];
int top;
bool cmp(node X, node Y){
    if(X.x != Y.x) return X.x < Y.x;
    if(X.y != Y.y) return X.y < Y.y;
    return X.point < Y.point;
}
int tot;
void BFS(int x){
    queue<int>q;
    q.push(x);
    while(!q.empty()){
        int u = q.front(); q.pop();
        for(int i = head[u]; ~i; i = edge[i].nex){
            int v = edge[i].to;
            if(dis[u] < dis[v])
            {
                dis[v] = dis[u];
                q.push(v);
            }
        }
    }
}
void input(){
    mp.clear();
    init();
    top = tot = 0;
    for(int i = 1; i <= n; i++)
    {
        cin>>s[i]; go(s[i]);
        if(mp.count(s[i])==0)
            {
                S[i] = mp[s[i]] = ++tot;
            int len = s[i].length(), num = 0;
            for(int j = 0; j < len; j++) num += s[i][j]=='r';
            st[top++] = node(num, len, tot);
            LEN[tot] = len; NUM[tot] = num;
        }
        else S[i] = mp[s[i]];
    }
    scanf("%d", &m);
    for(int i = 1; i <= m; i++)
    {
        cin>>a[i]>>b[i]; go(a[i]); go(b[i]);
        if(mp.count(a[i])==0)
        {
            A[i] = mp[a[i]] = ++tot;
            int len = a[i].length(), num = 0;
            for(int j = 0; j < len; j++) num += a[i][j]=='r';
            st[top++] = node(num, len, tot);
            LEN[tot] = len; NUM[tot] = num;
        }
        else A[i] = mp[a[i]];
        if(mp.count(b[i])==0)
        {
            B[i] = mp[b[i]] = ++tot;
            int len = b[i].length(), num = 0;
            for(int j = 0; j < len; j++) num += b[i][j]=='r';
            st[top++] = node(num, len, tot);
            LEN[tot] = len; NUM[tot] = num;
        }
        else B[i] = mp[b[i]];
        add(B[i], A[i]);
    }
}
int main(){
    while(~scanf("%d", &n)){
        input();
        for(int i = 1; i <= tot; i++)dis[i] = node(NUM[i], LEN[i],0);
        sort(st, st+top, cmp);
        for(int i = 0; i < top; i++)
            BFS(st[i].point);
        long long ans1 = 0, ans2 = 0;
        for(int i = 1; i <= n; i++)     
            ans1 += dis[S[i]].x, ans2 += dis[S[i]].y;
        
        printf("%I64d %I64d\n", ans1, ans2);
    }
    return 0;
}

Codeforces 467D Fedor and Essay bfs