[IOI2007]Miners 矿工配餐

link

其实就是一个比较简单的$IOI$题。简单$dp$就行，设$5$维$dp$即可

最后在滚动一下，判一下可行性即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
inline int read(){
    int f=,ans=;char c=getchar();
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){ans=ans*+c-'';c=getchar();}
    return f*ans;
}
const int N=;
int dp[N][][][][];
int val[],A[];
int calc(int a,int b,int c){
    A[]=a,A[]=b,A[]=c;sort(A+,A++);
    int res=(A[]!=);
    if(A[]!=A[]&&A[]) res++;
    if(A[]!=A[]&&A[]) res++;
    return res;
}
int maxn;
int cur;
int main(){
    cur=;
    memset(dp,-,sizeof(dp));dp[][][][][]=;
    int n=read();
    for(int i=;i<=n;i++){char str=getchar();if(str=='M') val[i]=;if(str=='F')val[i]=;if(str=='B') val[i]=;}
    for(int i=;i<=n;i++){
        cur^=;
        for(int t1=;t1<=;t1++)
            for(int t2=;t2<=;t2++)
                for(int t3=;t3<=;t3++)
                    for(int t4=;t4<=;t4++){
                        if(dp[cur^][t1][t2][t3][t4]==-) continue;
                        dp[cur][t2][val[i]][t3][t4]=max(dp[cur][t2][val[i]][t3][t4],dp[cur^][t1][t2][t3][t4]+calc(t1,t2,val[i]));
                        dp[cur][t1][t2][t4][val[i]]=max(dp[cur][t1][t2][t4][val[i]],dp[cur^][t1][t2][t3][t4]+calc(t3,t4,val[i]));
                    }
        memset(dp[cur^],-,sizeof(dp[cur^]));
    }
    for(int t1=;t1<=;t1++)
        for(int t2=;t2<=;t2++)
            for(int t3=;t3<=;t3++)
                for(int t4=;t4<=;t4++) if(dp[cur][t1][t2][t3][t4]!=-)maxn=max(maxn,dp[cur][t1][t2][t3][t4]);
    cout<<maxn;
}