Codeforces 467D - Fedor and Essay (反向建图 + DFS)

题意

给出原文的几个单词和几组替换词，要求最后包含的r最少，长度越短越好。

思路

因为替换的终点肯定可以确定（R最少的单词），所以可以反向建图，把所有单词按照r排序以后DFS一遍，把它们能走到的点的r和size都更新为自己的r、size。

然后直接统计文章中的单词的r和sz

代码

#include <stack>

#include <cstdio>

#include <list>

#include <set>

#include <iostream>

#include <string>

#include <vector>

#include <queue>

#include <functional>

#include <cstring>

#include <algorithm>

#include <cctype>

#include <string>

#include <map>

#include <cmath>

using namespace std;

#define LL long long

#define ULL unsigned long long

#define SZ(x) (int)x.size()

#define Lowbit(x) ((x) & (-x))

#define MP(a, b) make_pair(a, b)

#define MS(arr, num) memset(arr, num, sizeof(arr))

#define PB push_back

#define X first

#define Y second

#define ROP freopen("input.txt", "r", stdin);

#define MID(a, b) (a + ((b - a) >> 1))

#define LC rt << 1, l, mid

#define RC rt << 1|1, mid + 1, r

#define LRT rt << 1

#define RRT rt << 1|1

#define BitCount(x) __builtin_popcount(x)

#define BitCountll(x) __builtin_popcountll(x)

#define LeftPos(x) 32 - __builtin_clz(x) - 1

#define LeftPosll(x) 64 - __builtin_clzll(x) - 1

const double PI = acos(-1.0);

const int INF = 0x3f3f3f3f;

const double eps = 1e-8;

const int MAXN = 2e5 + 10;

const int MOD = 29;

const int dir[][2] = { {1, 0}, {0, 1} };

int cases = 0;

typedef pair<int, int> pii;

struct NODE

{

int id, rNum, sz;

bool operator < (const NODE &a) const

{

if (rNum != a.rNum) return rNum < a.rNum;

return sz < a.sz;

}

};

void ChangeToLower(string &str)

{

for (int i = 0; i < SZ(str); i++) str[i] = tolower(str[i]);

}

int CountR(const string &str)

{

int cnt = 0;

for (int i = 0; i < SZ(str); i++)

if (str[i] == 'r') cnt++;

return cnt;

}

map<string, int> oddId;

map<int, int> newId;

int cnt;

vector<NODE> words;

int GetNumber(string &tmp)

{

ChangeToLower(tmp);

if (oddId.count(tmp)) return oddId[tmp];

oddId[tmp] = cnt;

int r = CountR(tmp);

words.PB((NODE){cnt, r, SZ(tmp)});

cnt++;

return cnt - 1;

}

vector<int> G[MAXN], article;

int vis[MAXN];

void DFS(int curId, int r, int sz)

{

vis[curId] = 1;

for (int i = 0; i < SZ(G[curId]); i++)

{

int tmpId = G[curId][i];

if (!vis[tmpId])

{

vis[tmpId] = 1;

words[newId[tmpId]].rNum = r;

words[newId[tmpId]].sz = sz;

DFS(tmpId, r, sz);

}

}

}

int main()

{

//ROP;

int n;

cin >> n;

for (int i = 0; i < n; i++)

{

string tmp;

cin >> tmp;

int curNumber = GetNumber(tmp);

article.PB(curNumber);

}

int ncnv;

cin >> ncnv;

for (int i = 0; i < ncnv; i++)

{

string tmp;

cin >> tmp;

int u = GetNumber(tmp);

cin >> tmp;

int v = GetNumber(tmp);

G[v].PB(u);

}

sort(words.begin(), words.end());

for (int i = 0; i < SZ(words); i++) newId[words[i].id] = i;

for (int i = 0; i < SZ(words); i++)

if (!vis[words[i].id]) DFS(words[i].id, words[i].rNum, words[i].sz);

LL rAns = 0, szAns = 0;

for (int i = 0; i < SZ(article); i++)

rAns += words[newId[article[i]]].rNum, szAns += words[newId[article[i]]].sz;

cout << rAns << " " << szAns << endl;

return 0;

}