【POJ 1141 括号序列】 區間dp

POJ 1141

題意

就是空串是合法的

S為合法串

(S)合法

[S]合法

問你最少添加幾個字元串使得輸入的串為合法串并輸出合法串

我們用區間dp搞一下 dp[l][r]代表從 l 到 r 需要加幾個串

然後統計出答案用 dfs 跑一下序列 輸出答案

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 205;
char str[MAX_N];
int dp[MAX_N][MAX_N];
bool check(int l,int r)
{
    if((str[l]=='('&&str[r]==')')||(str[l]=='['&&str[r]==']')) return true;
    return false;
}
void dfs(int l,int r)
{
    if(l>r) return ;
    if(l==r)
    {
        if(str[l]=='('||str[l]==')') printf("()");
        if(str[l]=='['||str[l]==']') printf("[]");
        return;
    }
    int now = dp[l][r];
    if(now==dp[l+1][r-1]&&check(l,r))
    {
        printf("%c",str[l]);
        dfs(l+1,r-1);
        printf("%c",str[r]);
        return;
    }
    for(int k = l;k<r;++k)
    {
        if(now==dp[l][k]+dp[k+1][r])
        {
            dfs(l,k);dfs(k+1,r);return;
        }
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%s",str+1);
    int len = strlen(str+1);
    //memset(dp,0x3f,sizeof(dp));
    for(int i = 1;i<=len;++i) dp[i][i] = 1;
    for(int l = 1;l<len;++l)
    {
        for(int j = 1;j+l<=len;++j)
        {
            int r = j + l;
            dp[j][r] = 0x3f3f3f3f;
            if(check(j,r))
                dp[j][r] = min(dp[j+1][r-1],dp[j][r]);
            for(int k = j;k<r;++k)
                dp[j][r] = min(dp[j][r],dp[j][k]+dp[k+1][r]);
        }
    }
    dfs(1,len);
    printf("\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}