P1032 字串变换_BFS+判重

P1032 字串变换 传送门

不描述题目了.

应该说作为BFS题目还是不难的, 看着就有思路, 题目的数据也是比较小的了.

思路就是: BFS+模拟匹配+去重

注意, 去重是万万不可少的, 不然程序可能会运行很长的时间, 而去重可以大大减少时间.

还要注意不满足条件是NO ANSWER!, 这里有个感叹号.

#include <algorithm>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
using namespace std;

set<string> see;

string change(string now, pair<string, string> vs, int start)
{
    string pattern = vs.first, prod = vs.second;
    int i;
    for (i = start; i < start + pattern.size(); ++i) {
        if (now[i] != pattern[i - start]) return "";
    }
    string ret;
    for (int j = 0; j < start; ++j) {
        ret += now[j];
    }
    ret += prod;
    while (i != now.size()) {
        ret += now[i];
        i++;
    }
    return ret;
}

int bfs(vector<pair<string, string> > vs, string str1, string str2)
{
    queue<pair<string, int> > que;
    que.push(make_pair(str1, 0));
    while (!que.empty()) {
        string now = que.front().first;
        int time = que.front().second;
        //cout << "now = " << now << "  time = " << time << endl; 
        if (time > 10) return -1; 
        que.pop();
        if (now == str2) return time;
        for (int i = 0; i < vs.size(); ++i) {
            //cout << "i = " << i << "  vs[i].first = " << vs[i].first << endl; 
            if (now.size() < vs[i].first.size()) continue;
            for (int j = 0; j <= now.size() - vs[i].first.size(); ++j) {
                /*if (j == 1 && vs[i].first == "ud") {
                    cout << "匹配结果: " << change(now, vs[i], j) << endl; 
                }*/
                string next = change(now, vs[i], j);
                if (see.count(next)) continue;
                else see.insert(next);
                //cout << "next = " << next << endl;
                if (next == "") continue;
                que.push(make_pair(next, time + 1));
                //break;
                //cout << "push le " << next << endl;
            }
        }
    }
    return -1;
}

int main()
{
    string str1, str2, s1, s2;
    cin >> str1 >> str2;
    vector<pair<string, string> > vs;
    while(cin >> s1 >> s2) {
        vs.push_back(make_pair(s1, s2));
    }
    int ans = bfs(vs, str1, str2);
    if (~ans) cout << ans;
    else cout << "NO ANSWER!";
}
/*
abaaaba abcdaba
a b
b d
d e
e f
f g
g c // 8
*/
/*
abcd xyz
abc xu
ud y
y yz
*/ 
/*
BFS模拟遍历过程
找到可匹配的字符串,加入队列当中
不断匹配直到成功或次数达到限制
有一个坑就是一个字符串可能有多个可以匹配的段. 
*/