POJ 1179 Interval DP

https://vjudge.net/problem/POJ-1179

You should not only DP the Max

but also DP the min

And using the double length

the time will from the n^4 reduce to the n^3

顺便VJ上400道题

纪念一下

WildCow 学长他们队在大连站回放赛上AK了

强啊

龙哥和杨哥拿了个蓝桥杯国赛三等奖

好强啊

我好弱啊啊啊

入坑半年多了

感觉良好

因为如果我现在没有沉迷ACM

肯定已经沉溺于游戏了(大雾)

Code of AC:

#include<iostream>
using namespace std;
const int N=120;
const int inf=1<<30;
int A[N],B[N],DP[N][N][2];
int main(){
    int n;
    cin>>n;
    char op;
    for(int i=1;i<=n;++i){
        cin>>op;
        if(op=='x') B[i]=B[n+i]=1;
        cin>>A[i];
        A[n+i]=A[i];
    }
    for(int i=1;i<=2*n;++i) DP[i][i][1]=DP[i][i][0]=A[i];
    for(int i=1;i<2*n;++i){
        for(int j=1;j+i<=2*n;++j){
            DP[j][i+j][0]=inf;
            DP[j][i+j][1]=-inf;
            for(int k=j;k<i+j;++k){
                int Min=inf;
                int Max=-inf;
                if(B[k+1]){
                    Min=min(min(DP[j][k][0]*DP[k+1][i+j][0],DP[j][k][0]*DP[k+1][i+j][1]),
                        min(DP[j][k][1]*DP[k+1][i+j][0],DP[j][k][1]*DP[k+1][i+j][1]));
                    Max=max(max(DP[j][k][0]*DP[k+1][i+j][0],DP[j][k][0]*DP[k+1][i+j][1]),
                        max(DP[j][k][1]*DP[k+1][i+j][0],DP[j][k][1]*DP[k+1][i+j][1]));
                }
                else{
                    Min=min(min(DP[j][k][0]+DP[k+1][i+j][0],DP[j][k][0]+DP[k+1][i+j][1]),
                        min(DP[j][k][1]+DP[k+1][i+j][0],DP[j][k][1]+DP[k+1][i+j][1]));
                    Max=max(max(DP[j][k][0]+DP[k+1][i+j][0],DP[j][k][0]+DP[k+1][i+j][1]),
                        max(DP[j][k][1]+DP[k+1][i+j][0],DP[j][k][1]+DP[k+1][i+j][1]));
                }
                DP[j][i+j][0]=min(DP[j][i+j][0],Min);
                DP[j][i+j][1]=max(DP[j][i+j][1],Max);
            }
        }
    }
    int ans=-inf;
    for(int i=1;i<=n;++i) ans=max(ans,DP[i][i+n-1][1]);
    cout<<ans<<endl;
    int flag=0;
    for(int i=1;i<=n;++i){
        if(ans==DP[i][i+n-1][1]){
            if(flag) cout<<" ";
            flag=1;
            cout<<i;
        } 
    }
    cout<<endl;
}