紫书动规 例题9-10 UVA

题目链接:

https://vjudge.net/problem/UVA-1626

题意:

题解:

dp[i][j]:= i~j需要最少的括号
区间dp：
dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]);
if(match(s[i],s[j])) dp[i][j] = min(dp[i][j],dp[i+1][j-1]);

学到了一种dp打印解得方式

读入真坑啊…

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

char s[maxn];
int dp[105][105];

bool match(char s1,char s2){
    if((s1=='(' && s2==')') || (s1=='['&&s2==']')) return true;
    return false;
}

void print(int i,int j){
    if(i > j) return ;
    if(i == j){
        if(s[i]=='(' || s[i]==')') printf("()");
        else printf("[]");
        return ;
    }
    int ans = dp[i][j];
    if(match(s[i],s[j]) && ans==dp[i+1][j-1]) {
        printf("%c",s[i]); print(i+1,j-1); printf("%c",s[j]);
        return ;
    }

    for(int k=i; k<j; k++){
        if(ans == dp[i][k]+dp[k+1][j]){
            print(i,k); print(k+1,j);
            return ;
        }
    }
}

int main(){
    int T;
    gets(s);
    sscanf(s,"%d",&T);
    gets(s);

    while(T--){
        gets(s);
        int n = strlen(s);
        for(int i=0; i<n; i++){
            dp[i+1][i] = 0;
            dp[i][i] = 1;
        }

        for(int len=1; len<n; len++){
            for(int i=0; i+len<n; i++){
                int j = i+len;
                dp[i][j] = n;
                if(match(s[i],s[j])) dp[i][j] = min(dp[i][j],dp[i+1][j-1]);
                for(int k=i; k<j; k++){
                    dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }

        print(0,n-1);
        puts("");
        if(T) puts("");
        gets(s);
    }

    return 0;
}