Acwing P283 多边形 题解

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define int long long 
  6 #define maxn 50+10
  7 #define INF 9223372036854775807
  8 using namespace std;
  9 inline int read() 
 10 {
 11     int x=0;
 12     bool f=1;
 13     char c=getchar();
 14     for(; !isdigit(c); c=getchar()) if(c==‘-‘) f=0;
 15     for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+c-‘0‘;
 16     if(f) return x;
 17     return 0-x;
 18 }
 19 inline void write(int x)
 20 {
 21     if(x<0){putchar(‘-‘);x=-x;}
 22     if(x>9)write(x/10);
 23     putchar(x%10+‘0‘);
 24 }
 25 int n;
 26 int sym[2*maxn],num[2*maxn];
 27 int dp[maxn][2*maxn][2*maxn],mdp[maxn][2*maxn][2*maxn];
 28 int ans;
 29 inline int max_five(int a,int b,int c,int d,int e)
 30 {
 31     return max(max(max(a,b),max(c,d)),e);
 32 }
 33 inline int min_five(int a,int b,int c,int d,int e)
 34 {
 35     return min(min(min(a,b),min(c,d)),e);
 36 }
 37 inline void clear()
 38 {
 39     for(int i=1;i<=50;i++)
 40     {
 41         for(int j=1;j<=100;j++)
 42         {
 43             for(int k=1;k<=100;k++)
 44             {
 45                 dp[i][j][k]=-INF;
 46                 mdp[i][j][k]=INF;
 47             }
 48         }
 49     }
 50 }
 51 signed main()
 52 {
 53 //    freopen("ploygon.in","r",stdin);
 54 //    freopen("ploygon.out","w",stdout);
 55     clear();
 56     n=read();
 57     for(int i=1;i<=2*n;i++)
 58     {
 59         if(i%2==1) 
 60         {
 61             char in;
 62             cin>>in;
 63             if(in==‘t‘) sym[i/2+1]=1;
 64             else if(in==‘x‘) sym[i/2+1]=2;
 65         }
 66         else if(i%2==0) num[i/2]=read();
 67     }
 68     for(int i=n+1;i<=2*n;i++) 
 69     {
 70         sym[i]=sym[i-n];
 71         num[i]=num[i-n];
 72     }
 73     for(int j=1;j<=n;j++)
 74     {
 75         for(int i=1;i<=2*n;i++)
 76         {
 77             dp[j][i][i]=num[i];
 78             mdp[j][i][i]=num[i];
 79         }
 80     }
 81     for(int i=1;i<=n;i++) 
 82     {
 83         for(int len=2;len<=n;len++)
 84         {
 85             for(int j=i;j<=i+n-1;j++)
 86             {
 87                 int k=j+len-1;
 88                 if(k>=i+n) break;
 89                 for(int l=j;l<k;l++)
 90                 {
 91                     if(sym[l+1]==1)
 92                     {
 93                         dp[i][j][k]=max(dp[i][j][k],dp[i][j][l]+dp[i][l+1][k]);
 94                         mdp[i][j][k]=min(mdp[i][j][k],mdp[i][j][l]+mdp[i][l+1][k]);
 95                     } 
 96                     else if(sym[l+1]==2) 
 97                     {
 98                         dp[i][j][k]=max_five(dp[i][j][k],dp[i][j][l]*dp[i][l+1][k],mdp[i][j][l]*dp[i][l+1][k],dp[i][j][l]*mdp[i][l+1][k],mdp[i][j][l]*mdp[i][l+1][k]);
 99                         mdp[i][j][k]=min_five(dp[i][j][k],dp[i][j][l]*dp[i][l+1][k],mdp[i][j][l]*dp[i][l+1][k],dp[i][j][l]*mdp[i][l+1][k],mdp[i][j][l]*mdp[i][l+1][k]);
100                     }
101                 }
102             }
103         }
104         ans=max(ans,dp[i][i][i+n-1]);
105     }
106     write(ans);
107     printf("\n");
108     for(int k=1;k<=n;k++)
109     {
110         if(dp[k][k][k+n-1]==ans)
111         {
112             write(k);
113             printf(" ");
114         }
115     }
116     return 0;
117 }