【cf1272】F. Two Bracket Sequences
作者：互联网
题意：
给出\(s,t\)两个合法括号序列，现在找到一个长度最小的合法的序列\(p\)，使得\(s,t\)都为其子序列。
思路：
- 考虑\(dp:dp[i][j][d]\)表示第一个串在\(i\)，第二个串在\(j\)，答案串左括号和右括号之差为\(d\)时的最短长度。
- 那么转移时枚举下一位转移即可。
- 还需要考虑一点细节：若当前\(d=0\)，但是下一位为)时，我们需要先\(dp[i][j][0]\)向\(dp[i][j][1]\)转移，即在答案串中添加一个(来匹配。
- 最后的答案即为\(min\{dp[n][m][d]+d\}\)，若\(d>0\)时，我们还需要在答案串后面添加\(d\)个)来使其合法。
- 输出路径时转移时记录一下前驱，然后根据\(d\)来找到当前应为哪个括号。
代码如下：
/*
* Author: heyuhhh
* Created Time: 2019/12/13 11:40:50
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << '\n'; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
#define dbg(...)
#endif
void pt() {std::cout << '\n'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 200 + 5;
struct node {
int i, j, k;
};
node pre[N][N][N << 1];
int dp[N][N][N << 1];
int n, m;
char s[N], t[N];
void run(){
cin >> (s + 1) >> (t + 1);
n = strlen(s + 1);
m = strlen(t + 1);
memset(dp, INF, sizeof(dp));
dp[0][0][0] = 0;
for(int i = 0; i <= n; i++) {
for(int j = 0; j <= m; j++) {
if(i == n && j == m) break;
for(int k = 0; k < N; k++) if(dp[i][j][k] != INF) {
if(s[i + 1] == t[j + 1]) {
if(s[i + 1] == ')') {
if(k == 0) {
if(dp[i][j][k + 1] > dp[i][j][k] + 1) {
dp[i][j][k + 1] = dp[i][j][k] + 1;
pre[i][j][k + 1] = node {i, j, k};
}
}
if(dp[i + 1][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k - 1] = dp[i][j][k] + 1;
pre[i + 1][j + 1][k - 1] = node {i, j, k};
}
} else if(s[i + 1] == '(') {
if(dp[i + 1][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k + 1] = dp[i][j][k] + 1;
pre[i + 1][j + 1][k + 1] = node {i, j, k};
}
}
} else {
if(s[i + 1] == ')') {
if(k == 0) {
if(dp[i][j][k + 1] > dp[i][j][k] + 1) {
dp[i][j][k + 1] = dp[i][j][k] + 1;
pre[i][j][k + 1] = node {i, j, k};
}
}
if(dp[i + 1][j][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k - 1] = dp[i][j][k] + 1;
pre[i + 1][j][k - 1] = node {i, j, k};
}
} else if(s[i + 1] == '(') {
if(dp[i + 1][j][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k + 1] = dp[i][j][k] + 1;
pre[i + 1][j][k + 1] = node {i, j, k};
}
}
if(t[j + 1] == ')') {
if(k == 0) {
if(dp[i][j][k + 1] > dp[i][j][k] + 1) {
dp[i][j][k + 1] = dp[i][j][k] + 1;
pre[i][j][k + 1] = node {i, j, k};
}
}
if(dp[i][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k - 1] = dp[i][j][k] + 1;
pre[i][j + 1][k - 1] = node {i, j, k};
}
} else if(t[j + 1] == '(') {
if(dp[i][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k + 1] = dp[i][j][k] + 1;
pre[i][j + 1][k + 1] = node {i, j, k};
}
}
}
}
}
}
int Min = INF;
node ans;
for(int i = 0; i < N; i++) {
if(dp[n][m][i] + i < Min) {
Min = dp[n][m][i] + i;
ans = node {n, m, i};
}
}
//for(int i = 400; i < 410; i++) {
//cout << dp[1][3][i] << '\n';
//}
string res = "";
node now = ans, last;
while(1) {
last = pre[now.i][now.j][now.k];
//dbg(now.i, now.j, now.k, dp[now.i][now.j][now.k]);
if(last.k > now.k) res += ")";
else res += "(";
now = last;
if(now.i == 0 && now.j == 0 && now.k == 0) break;
}
reverse(res.begin(), res.end());
for(int i = 0; i < ans.k; i++) res += ")";
cout << res << '\n';;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}标签：node,pre,include,int,Two,Bracket,cf1272,dp,define 来源： https://www.cnblogs.com/heyuhhh/p/12043634.html