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poj2955(区间dp)

 

题目连接:http://poj.org/problem?id=2955

题意:给一个由()[]四种字符任意排列组成的字符串,求最长合法的不连续字串的长度。

分析:如果找到一对匹配的括号[xxx]oooo,就把区间分成两部分,一部分是xxx,一部分是ooo,然后以此递归直到区间长度为<=1. 

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#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <iostream>#include <algorithm>#include <queue>#include <cstdlib>#include <stack>#include <vector>#include <set>#include <map>#define LL long long#define mod 100000000#define inf 0x3f3f3f3f#define eps 1e-9#define N 100010#define FILL(a,b) (memset(a,b,sizeof(a)))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int dp[110][110];char str[110];bool judge(int i,int j){    return (str[i]==(&&str[j]==))||(str[i]==[&&str[j]==]);}int dfs(int l,int r){    if(dp[l][r]!=-1)return dp[l][r];    if(l>=r)return 0;    int temp=dfs(l+1,r);    for(int i=l+1;i<=r;i++)    {        if(judge(l,i))            temp=max(temp,dfs(l+1,i-1)+dfs(i+1,r)+2);    }    return dp[l][r]=temp;}int main(){    int n;    while(scanf("%s",str+1)>0)    {        if(strcmp(str+1,"end")==0)break;        FILL(dp,-1);n=strlen(str+1);        printf("%d\n",dfs(1,n));    }}
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poj2955(区间dp)