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POJ 1141 Brackets Sequence (区间DP)

题意:给定一个括号序列,让你添加最少的括号,使得所有的括号都匹配。

析:首先用DP来把这个最少的找出来,然后再打印出解,dp[i][j]表示从 i 到 j 所要添加最少的数。

注意有空行的数据。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 100 + 5;const int mod = 1e9 + 7;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}char s[maxn];int dp[maxn][maxn];bool judge(int i, int j){    if(s[i] == ‘(‘ && s[j] == ‘)‘)  return true;    if(s[i] == ‘[‘ && s[j] == ‘]‘)  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(judge(i, 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(){    while(gets(s) != NULL){        n = strlen(s);        memset(dp, INF, sizeof dp);        for(int i = 0; i < n; ++i)  dp[i][i] = 1, dp[i+1][i] = 0;        for(int i = n-2; i >= 0; --i)            for(int j= i+1; j < n; ++j){                dp[i][j] = INF;                if(judge(i, j))  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);        printf("\n");    }    return 0;}

 

POJ 1141 Brackets Sequence (区间DP)