luogu P4342 [IOI1998]Polygon

IOI早期这么多dp？

题目要求断掉环上的一边,我们可以断环为链,开两倍数组

容易想到dp,设\(f_{i,j}\)为区间\([i,j]\)的最大值,然后就是个枚举断点的区间dp

不过可能会有负数出现,这意味着可能区间中可能会有两个负数相乘得到最大值的情况,所以设\(g_{i,j}\)为区间\([i,j]\)的最小值

转移时记得考虑所有可能情况

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define LL long long
#define il inline
#define re register
#define db double
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

using namespace std;
const int N=100+10;
const LL inf=1e14;
il LL rd()
{
    re LL x=0,w=1;re char ch=0;
    while(ch<'0'||ch>'9') {if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    return x*w;
}
LL f[N][N],g[N][N],ans=-inf,an[N],a[N][2],tt;
int n,m;

int main()
{
  n=rd();
  for(int i=1;i<=n;i++)
  {
  	char cc[2];
  	scanf("%s",cc);
  	a[i][1]=a[i+n][1]=(cc[0]=='t');
  	a[i][0]=a[i+n][0]=f[i][i]=f[i+n][i+n]=g[i][i]=g[i+n][i+n]=rd();
  }
  for(int l=1;l<=n-1;l++)
  	for(int i=1,j=i+l;j<(n<<1);i++,j++)
  	{
  	  f[i][j]=-inf,g[i][j]=inf;
  	  for(int k=i+1;k<=j;k++)
  	  {
  	  	if(a[k][1])
  	  	  f[i][j]=max(f[i][j],f[i][k-1]+f[k][j]),
  	  	  g[i][j]=min(g[i][j],g[i][k-1]+g[k][j]);
  	  	else
  	  	  f[i][j]=max(f[i][j],max(f[i][k-1]*f[k][j],g[i][k-1]*g[k][j])),
  	  	  g[i][j]=min(g[i][j],min(f[i][k-1]*f[k][j],g[i][k-1]*g[k][j])),
  	  	  g[i][j]=min(g[i][j],min(f[i][k-1]*g[k][j],g[i][k-1]*f[k][j]));
  	  }
  	  //printf("%d %d %lld %lld\n",i,j,f[i][j],g[i][j]);
  	}
  for(int i=1;i<=n;i++)
  	if(ans<f[i][i+n-1]) ans=f[i][i+n-1],an[tt=1]=i;
  	else if(ans==f[i][i+n-1]) an[++tt]=i;
  printf("%lld\n",ans);
  for(int i=1;i<=tt;i++) printf("%lld ",an[i]);
  return 0;
}