輸出路徑的過程
結合dp數組和代碼一看就懂了
a b d k s c a b
a 1 1 1 1 1 1 1 1
b 1 2 2 2 2 2 2 2
c 1 2 2 2 2 3 3 3
i 1 2 2 2 2 3 3 3
c 1 2 2 2 2 3 3 3
b 1 2 2 2 2 3 3 4
a 1 2 2 2 2 3 4 4
經典的dp題,關鍵題目讓輸出最後的公共序列,可以通過判斷dp數組來從後往前找到狀态變化的字母,最後反序輸出
狀态轉移方程:
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define d(x) cout << (x) << endl
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
const int N = 1e3 + 10;
const int M = 1e4 + 10;
int dp[N][N];
int main()
{
string a, b, c = "";
cin >> a >> b;
for (int i = 1; i <= a.size(); i++) {
for (int j = 1; j <= b.size(); j++) {
if(a[i-1] == b[j-1]){
dp[i][j] = dp[i - 1][j - 1] + 1;
}else{
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
int i = a.size() - 1;
int j = b.size() - 1;
while(i >= 0 && j >= 0){
if(a[i] == b[j]){
c += a[i];
i--;
j--;
}else if(dp[i][j+1] > dp[i+1][j]){
i--;
}else{
j--;
}
}
reverse(c.begin(), c.end());
cout << c << endl;
return 0;
}