微软2014编程之美初赛第二场——题目2 : 字符串压缩

【来源】

题目2 : 字符串压缩

【分析】

把游程编码恢复为原始字符串，然后得出每一行的字符串的内容，放在一个vector中。用map统计vector中每一行的反复的次数。

比較两个游程编码得到的map是否同样就可以。

该算法占用空间太多，小数据AC，大数据MLE了。

【代码】

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <sstream>
using namespace std;

int main()
{
    int T;
    cin >> T;
    for (int c = 0; c < T; ++c){
        int L;
        cin >> L;
        string code1, code2;
        cin >> code1 >> code2;
        stringstream ss1;
        ss1 << code1 + "0A";
        stringstream ss2;
        ss2 << code2 + "0A";

        string raw1 = "";
        string raw2 = "";
        while(!ss1.eof()){
            int freq;
            ss1 >> freq;
            char c;
            ss1 >> c;
            if (freq == 0){
                break;
            }
            for (int j = 0; j < freq; ++j){
                raw1 += c;
            }
        }

        while (!ss2.eof()){
            int freq;
            ss2 >> freq;
            char c;
            ss2 >> c;
            if (freq == 0){
                break;
            }
            for (int j = 0; j < freq; ++j){
                raw2 += c;
            }
        }

        int lines = raw1.size()/L;
        vector<string> lines1, lines2;
        map<string, int> map1, map2;
        for (int i = 0; i < lines; ++i){
            string line = "";
            for (int j = 0; j < L; ++j){
                line += raw1[i+j*lines];
            }
            if (map1[line] == 0){
                map1[line] = 1;
            }
            else{
                ++map1[line];
            }
            lines1.push_back(line);
        }

        for (int i = 0; i < lines; ++i){
            string line = "";
            for (int j = 0; j < L; ++j){
                line += raw2[i + j*lines];
            }
            if (map2[line] == 0){
                map2[line] = 1;
            }
            else{
                ++map2[line];
            }
            lines2.push_back(line);
        }
        int ii;
        for (ii = 0; ii < lines; ++ii){
            if (map1[lines1[ii]] != map2[lines1[ii]]){

                break;
            }
        }
        int jj;
        for (jj = 0; jj < lines; ++jj){
            if (map1[lines2[jj]] != map2[lines2[jj]]){

                break;
            }
        }

        if ((ii == lines) && (jj == lines) && (raw1.size() == raw2.size())){
            cout << "Case " << c + 1 << ": Yes" << endl;
        }
        else{
            cout << "Case " << c + 1 << ": No" << endl;
        }
    }

    //system("pause");
    return 0;
}

【点评】

字符串处理题。当时想的算法太占用空间了，应该能够利用文本内容仅仅有a-z这26个字符来做文章，比方Hash什么的。