【DP】 POJ 2955 Brackets 区间DP

来源：互联网发布：埃舍尔知乎编辑：程序博客网时间：2024/05/22 08:08

从小区间到大大区间

并计算当前能否括号匹配

#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <iostream>#include <algorithm>#include <sstream>#include <cmath>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#include <map>#define cler(arr, val)    memset(arr, val, sizeof(arr))typedef long long  LL;const int MAXN = 10000+6;const int MAXM = 140000;const int INF = 0x3f3f3f3f;const int mod = 1000000007;char s[122],c[122];int dp[123][123],n;int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    // freopen("out.txt", "w", stdout);#endif    while(scanf("%s",s))    {        if(strcmp(s,"end")==0) break;        int len=strlen(s);        for(int i=0;i<len;i++)            c[i+1]=s[i];        cler(dp,0);        for(int i=1;i<len;i++)            if((c[i]=='('&&c[i+1]==')')||(c[i]=='['&&c[i+1]==']'))                dp[i][i+1]=2;        for(int i=1;i<=len;i++)//区间长度        {            for(int l=1;l+i<=len;l++)//起始位置            {                int r=l+i;                if((c[l]=='('&&c[r]==')')||(c[l]=='['&&c[r]==']'))//当前格子满足                    dp[l][r]=max(dp[l][r],dp[l+1][r-1]+2);                for(int k=l+1;k<=r-1;k++)//合并两个区间                {                    dp[l][r]=max(dp[l][r],dp[l][k]+dp[k][r]);                }            }        }        cout<<dp[1][len]<<endl;    }    return 0;}

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