icpc2018焦作Mathematical Curse（动态规划）

传送门

ac代码：

#include<bits/stdc++.h>
#define per(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
//#define int long long
const ll inf =2333333333333333LL;
const double eps=1e-8;
int read(){
    char ch=getchar();
    int res=0,f=0;
    while(ch<'0' || ch>'9'){f=(ch=='-'?-1:1);ch=getchar();}
    while(ch>='0'&&ch<='9'){res=res*10+(ch-'0');ch=getchar();}
    return res*f;
}
// ------------------------head
#define mod 1000000009
const int siz=10005;
ll dp[1005][6],k;
int T,n,m,a[1005];
char ch[6];
ll fun(int rhs,ll d,char op){
    if(op=='+')return d+(ll)rhs;
    else if(op=='-')return d-(ll)rhs;
    else if(op=='*')return d*(ll)rhs;
    else if(op=='/')return d/(ll)rhs; 
}

signed main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%lld",&n,&m,&k);
        ll ans=-inf;
        per(i,1,n)scanf("%d",&a[i]);
        per(i,1,m)cin>>ch[i];
        ll _mi[1005][6];
        per(i,0,n)dp[i][0]=_mi[i][0]=k;
        per(i,1,m){per(j,0,n)_mi[j][i]=inf;per(j,0,n)dp[j][i]=-inf;}
        per(i,1,n){
            dp[i][1]=max(dp[i-1][1],fun(a[i],k,ch[1]));
            _mi[i][1]=min(_mi[i-1][1],fun(a[i],k,ch[1]));
        }
        per(i,2,m){//区分正数和负数
            per(j,i,n){
                if(a[j]<0&&(ch[i]=='*'||ch[i]=='/')){
                    dp[j][i]=max(dp[j-1][i],fun(a[j],_mi[j-1][i-1],ch[i]));
                    _mi[j][i]=min(_mi[j-1][i],fun(a[j],dp[j-1][i-1],ch[i]));
                }
                else {
                    ll ma=dp[j-1][i-1];
                    dp[j][i]=max(dp[j-1][i],fun(a[j],dp[j-1][i-1],ch[i]));
                    _mi[j][i]=min(_mi[j-1][i],fun(a[j],_mi[j-1][i-1],ch[i]));
                }
            }
        }
        if(m==1){
            ll res=-inf;
            per(i,1,n)res=max(res,dp[i][1]);
            printf("%lld
",res);
        }
        else printf("%lld
",dp[n][m]);
    }
    return 0;
}