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  • poj1179

    题目的意思就是给n个数,n个两两数之间的运算符(只有+和*)问首先去掉哪个运算符号之后可以让其他的数按照一定的方法计算后结果最大。 其实结题思路还是比较好想到的,枚举(枚举去掉的符号)+DP(记忆化搜索)就可以做到。但这里有一个天坑,就是负负得正,所以不能单一的枚举最大值,而要同时DP最小值。        计算最大值:    加法  max(i,j) = max(i,k)+max(k,j);    乘法  max(i,j) = MAX(max(i,k)*max(k,j),max(i,k)*min(k,j),max(k,j)*min(i,k),min(i,k)*min(k,j));(i=<k<=j) 计算最小值:

       加法  min(i,j) = min(i,k)+min(k,j);    乘法  min(i,j) = MIN(max(i,k)*max(k,j),min(i,k)*min(k,j),max(k,j)*min(i,k),min(i,k)*min(k,j));(i=<k<=j)

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <stack>
    #include <set>
    #include <queue>
    #define MAX(a,b) (a) > (b)? (a):(b)
    #define MIN(a,b) (a) < (b)? (a):(b)
    #define mem(a) memset(a,0,sizeof(a))
    #define INF 1000000007
    #define MAXN 20005
    using namespace std;
    
    bool op[105];
    int num[105],dp_max[10005], dp_min[10005], n;
    bool vis_max[10005],vis_min[10005];
    int DP_MIN(int i,int j);
    int DP_MAX(int i,int j);
    
    
    int DP_MAX(int i,int j)//DP求区间最大值
    {
        int u = i*100+j;
        if(vis_max[u])return dp_max[u];
        vis_max[u]=1;
        if(j-i <= 1)
        {
            if(j==i)return dp_max[u]=num[i-1];
            if(!op[i])return dp_max[u]=num[i-1]+num[i];
            else return dp_max[u]=num[i-1]*num[i];
        }
        dp_max[u] = -INF;
        for(int k=i;k<j;k++)
        {
            int l=DP_MIN(i,k);
            int r=DP_MIN(k+1,j);
            int ll=DP_MAX(i,k);
            int rr=DP_MAX(k+1,j);
            if(!op[k])dp_max[u] = MAX(dp_max[u], ll+rr);
            else dp_max[u] = MAX(dp_max[u], MAX(ll*rr,MAX(l*r,MAX(l*rr,r*ll))));
        }
        return dp_max[u];
    }
    
    int DP_MIN(int i,int j)//DP求区间最小值
    {
        int u = i*100+j;
        if(vis_min[u])return dp_min[u];
        vis_min[u]=1;
        if(j-i <= 1)
        {
            if(j==i)return dp_min[u]=num[i-1];
            if(!op[i])return dp_min[u]=num[i-1]+num[i];
            else return dp_min[u]=num[i-1]*num[i];
        }
        dp_min[u] = INF;
        for(int k=i;k<j;k++)
        {
            int l=DP_MIN(i,k);
            int r=DP_MIN(k+1,j);
            int ll=DP_MAX(i,k);
            int rr=DP_MAX(k+1,j);
            if(!op[k])dp_min[u] = MIN(dp_min[u], l+r);
            else dp_min[u] = MIN(dp_min[u], MIN(ll*rr,MIN(l*r,MIN(l*rr,r*ll))));
        }
        return dp_min[u];
    }
    
    int main()
    {
        while(~scanf("%d%*c",&n))
        {
            mem(op);mem(dp_max);
            mem(num);mem(vis_min);
            mem(vis_max);
            int max=-INF,i;
            char ch;
            for(i=0;i<n;i++)
            {
                scanf("%c %d%*c",&ch,&num[i]);
                op[i]=op[i+n]=(ch=='x');
                num[i+n]=num[i];
            }
            for(i=0;i<n;i++)
            {
                max=MAX(max,DP_MAX(i+1,i+n));
            }
            printf("%d
    ",max);
            int ok=1;
            for(i=0;i<n;i++)
            {
                if(DP_MAX(i+1,i+n) == max)
                {
                    if(ok){printf("%d",i+1);ok=0;}
                    else printf(" %d",i+1);
                }
            }
            printf("
    ");
        }
        return 0;
    }
    View Code
    
    
    




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  • 原文地址:https://www.cnblogs.com/Kong-Ruo/p/7111647.html
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