題目連結

題目大意

  給兩個不同的字串，比較其相似度，你可以在字元之間插入'-'，不同字元之間的相似度參考題目中的表格。

解題思路

  這題主要還是考驗對LCS的理解。定義dp[i][j]表示第一個串s1長度為i時與第二個串s2長度為j時的相似度(注意不算'-')。
  我們在求dp[i][j]的時候，分三種情況：
  1.s1[i]與s2[j]相匹配，那麼結果就是直接長度為i-1的串s1與長度為j-1的串s2的匹配結果加上這兩個字元的匹配度。
  2.如果s1[i]與s2[j]之前的某個字元匹配了，那麼s2[j]就只能和'-'匹配了。
  3.如果s1[i]之前的某個字元與s2[j]的某個字元匹配了，那麼s1[i]就只能和'-'匹配了。
  dp的狀態轉移方程與LCS相似，注意dp[i][0]與dp[0][j]的初值，具體看程式碼吧。

程式碼

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<sstream>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define clr(arr,b) memset(arr, b, sizeof(arr))
#define IOS ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, int> Pll;
const double pi = acos(-1.0);
const double eps = 1e-8;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2+10;
const int maxm = 2e2+10;
int l1, l2, dp[maxn][maxn], mp[maxn][maxn], skp[maxn]; 
char s1[maxn], s2[maxn];
int main() {
    mp['A']['A'] = mp['C']['C'] = mp['G']['G'] = mp['T']['T'] = 5;
    mp['A']['C'] = mp['C']['A'] = -1;
    mp['A']['G'] = mp['G']['A'] = -2;
    mp['A']['T'] = mp['T']['A'] = -1;
    mp['C']['G'] = mp['G']['C'] = -3;
    mp['C']['T'] = mp['T']['C'] = -2;
    mp['G']['T'] = mp['T']['G'] = -2;
    skp['A'] = -3; skp['C'] = -4, skp['G'] = -2, skp['T'] = -1;
    int t; cin >> t;
    while(t--) {
        cin >> l1 >> (s1+1) >> l2 >> (s2+1);
        for (int i = 1; i<=l1; ++i) dp[i][0] = dp[i-1][0]+skp[s1[i]];
        for (int i = 1; i<=l2; ++i) dp[0][i] = dp[0][i-1]+skp[s2[i]];
        for (int i = 1; i<=l1; ++i)
            for (int j = 1; j<=l2; ++j)
                dp[i][j] = max(dp[i-1][j-1]+mp[s1[i]][s2[j]], 
                max(dp[i-1][j]+skp[s1[i]], dp[i][j-1]+skp[s2[j]]));
        cout << dp[l1][l2] << endl;
    }
    return 0;
}