UVALive 7454 Parentheses

来源：互联网发布：剑侠情缘mac版编辑：程序博客网时间：2024/05/20 17:41

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5476

题意：有一个含'(' 和')'的字符串，每次操作可以将左括号变为右括号或者右括号变为左括号。问至少要变多少次才可以使得序列括号匹配。

思路：dp[i][j]表示[i,j]变为括号匹配的序列最小操作次数。dp[i][j]可以由dp[i+1][j-1]转移过来，或者枚举中间断开点由dp[i][k] + dp[k+1][j]转移过来。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 100000007const int maxn = 109;LL f[maxn][maxn];char s[maxn];int T;int main(){    cin>>T;    getchar();    while(T--)    {        int len;        scanf("%d",&len);        getchar();        gets(s+1);        Clean(f,0);        rep(L,1,len)        {            rep(i,1,len-L+1)            {                int j = i + L - 1;                f[i][j] = inf;                if ( i == j ) continue;                f[i][j] = min( f[i][j] , f[i+1][j-1] + ( s[i]=='('?0:1 ) + ( s[j]==')'?0:1 ) );                rep(k,i,j-1)                f[i][j] = min( f[i][j] , f[i][k] + f[k+1][j] );            }        }        printf("%lld\n",f[1][len]);    }    return 0;}

0 0