IOI 98 (POJ 1179)Polygon(区间DP)

很容易想到枚举第一步切掉的边，然后再计算能够产生的最大值。

联想到区间DP，令dp[i][l][r]为第一步切掉第i条边后从第i个顶点起区间[l,r]能够生成的最大值是多少。

但是状态不好转移，因为操作的符号不仅有‘+’，还有‘*’，加法的话，父区间的最大值显然可以从子区间的最大值相加得出。

乘法的话，父区间的最大值除了由子区间的最大值相乘得出，还可以由子区间的最小值相乘得出。

所以，多定义一维状态。 dp[i][l][r][flag]表示第一步切掉第i条边后从第i个顶点起区间[l,r]能够生成的最大值/最小值是多少？

转移的话很简单

dp[i][l][r][0]=min(dp[i][l][k][0]+dp[i][k+1][r][0])(操作符为+),min(dp[i][l][k][0]*dp[i][k+1][r][1], dp[i][l][k][1]*dp[i][k+1][r][0])(操作符为*).

dp[i][l][r][1]=max(dp[i][l][k][1]+dp[i][k+1][r][1])(操作符为+),min(dp[i][l][k][0]*dp[i][k+1][r][0], dp[i][l][k][1]*dp[i][k+1][r][1])(操作符为*).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=55;
//Code begin...

int ans[N], num[N], dp[N][N][N][2], n;
char s[N][2];
bool vis[N][N][N][2];

int dfs(int x, int l, int r, int flag)
{
    if (vis[x][l][r][flag]) return dp[x][l][r][flag];
    vis[x][l][r][flag]=1;
    if (l==r) return dp[x][l][r][flag]=num[(x+l-2)%n+1];
    if (flag) {
        int ans=-INF;
        FO(i,l,r) {
            if (s[(x+i-1)%n+1][0]=='t') ans=max(ans,dfs(x,l,i,1)+dfs(x,i+1,r,1));
            else ans=max(ans,max(dfs(x,l,i,1)*dfs(x,i+1,r,1),dfs(x,l,i,0)*dfs(x,i+1,r,0)));
        }
        return dp[x][l][r][flag]=ans;
    }
    else {
        int ans=INF;
        FO(i,l,r) {
            if (s[(x+i-1)%n+1][0]=='t') ans=min(ans,dfs(x,l,i,0)+dfs(x,i+1,r,0));
            else ans=min(ans,min(dfs(x,l,i,0)*dfs(x,i+1,r,1),dfs(x,l,i,1)*dfs(x,i+1,r,0)));
        }
        return dp[x][l][r][flag]=ans;
    }
}
int main ()
{
    int ma=-INF, flag=1;
    scanf("%d",&n);
    FOR(i,1,n) scanf("%s%d",s[i],num+i);
    FOR(i,1,n) ma=max(ma,dfs(i,1,n,1));
    printf("%d\n",ma);
    FOR(i,1,n) if (dp[i][1][n][1]==ma) printf(flag?"%d":" %d",i), flag=0;
    putchar('\n');
    return 0;
}