[IOI 1998] Polygon

[题目链接]

         http://poj.org/problem?id=1179

[算法]

        区间DP

[代码]

        

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;
#define MAXN 55
const long long INF = 1e15;

int i,j,k,l,r,n;
long long num[MAXN<<1];
long long mn[MAXN<<1][MAXN<<1],mx[MAXN<<1][MAXN<<1];
long long ans;
char op;
char opt[MAXN<<1];

int main()
{
    scanf("%d",&n);
    getchar();
    for (i = 1; i <= n; i++)
    {
        if (i > 1) scanf(" %c",&op);
        else scanf("%c",&op);
        scanf("%lld",&num[i]);
        opt[i] = opt[i + n] = (op == 't');
        num[i+n] = num[i];
    }
    for (i = 1; i <= 2 * n; i++)
    {
        for (j = i + 1; j <= 2 * n; j++)
        {
            mx[i][j] = -INF;
            mn[i][j] = INF;
        }
    }
    for (i = 1; i <= 2 * n; i++) mx[i][i] = mn[i][i] = num[i];
    for (i = 2; i <= n; i++)
    {
        for (l = 1; l <= 2 * n - i + 1; l++)
        {
            r = l + i - 1;
            for (k = l; k < r; k++)
            {
                if (opt[k+1])
                {
                    mn[l][r] = min(mn[l][r],mn[l][k] + mn[k+1][r]);
                    mx[l][r] = max(mx[l][r],mx[l][k] + mx[k+1][r]);
                } else
                {
                    mn[l][r] = min(mn[l][r],min(mx[l][k] * mx[k+1][r],(mx[l][k] * mn[k+1][r],mn[l][k] * mx[k+1][r])));
                    mx[l][r] = max(mx[l][r],max(mx[l][k] * mx[k+1][r],mn[l][k] * mn[k+1][r]));
                }
            }
        }
    }
    for (i = 1; i <= n; i++) ans = max(ans,mx[i][i+n-1]);
    printf("%lld
",ans);
    for (i = 1; i <= n; i++) 
    {
        if (mx[i][i+n-1] == ans)
            printf("%d ",i);
    }
    printf("
");

    return 0;
}