POJ - 2250 Compromise 最长公共子序列 LCS

题意：

给定两个字符串序列，求最长公共子序列

思路：

方便起见，把字符串映射成两个数列，求最长公共子系列

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <string>

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 100 + 7;

int a[maxn], b[maxn];
map<string, int> mp;
map<int, string> mpp;
string s;

int main() {
    int id = 1;
    int pos1 = 1, pos2 = 1;
    while(cin >> s) {
        mp[s] = id;
        mpp[id] = s;
        id++;
        a[pos1++] = mp[s];


        while(1) {
            cin >> s;
            if(s[0] == '#')
                break;
            if(mp[s])
                a[pos1++] = mp[s];
            else {
                mp[s] = id;
                mpp[id] = s;
                id++;
                a[pos1++] = mp[s];
            }
        }
        while(1) {
            cin >> s;
            if(s[0] == '#')
                break;
            if(mp[s])
                b[pos2++] = mp[s];
            else {
                mp[s] = id;
                mpp[id] = s;
                id++;
                b[pos2++] = mp[s];
            }
        }
        int dp[maxn][maxn] = {0};
        int d[maxn][maxn] = {0};
        for(int i = 1; i < pos1; ++i) {
            for(int j = 1; j < pos2; ++j) {
                if(a[i] == b[j]) {
                    dp[i][j] = max(dp[i-1][j-1]+1, max(dp[i][j-1], dp[i-1][j]));
                } else
                    dp[i][j] = max(dp[i-1][j-1], max(dp[i-1][j], dp[i][j-1]));
            }
        }
        int anss = dp[pos1-1][pos2-1];
        stack<int> sk;
        for(int i = pos1-1; i > 0; --i) {
            for(int j = pos2-1; j > 0; --j) {
                if(a[i] == b[j] && dp[i][j] == anss) {
                    anss--;
                    sk.push(a[i]);
                }
            }
        }
        while(!sk.empty()) {
            int tt = sk.top();
            cout << mpp[tt];
            sk.pop();
            if(sk.empty())
                cout << endl;
            else
                cout << " ";
        }
        id = 1;
        pos1 = 1;
        pos2 = 1;
        mp.clear();
    }

    return 0;
}