poj1179 Polygon（區間DP）

傳送門
這是一道環形DP，解決方法是“任意選擇一個位置斷開，複製形成2倍長度的鏈”。

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const ll INF = 1e18;

int n;
char opt[105];
ll num[105];
ll dp[105][105][2]; //dp[l][r][0], dp[l][r][1]分別表示區間最小，區間最大

int main(){
    int n;
    while(cin >> n){
        for(int i = 1; i <= n; ++i){
            cin >> opt[i-1];
            cin >> num[i];
        }
        opt[n] = opt[0];
        for(int i = n+1; i <= 2*n; ++i){
            num[i] = num[i-n];
            opt[i] = opt[i-n];
        }

        for(int i = 1; i <= 100; ++i)
            for(int j = i; j <= 100; ++j){
                dp[i][j][0] = INF;
                dp[i][j][1] = -INF;
            }
        for(int i = 1; i <= 100; ++i){
            dp[i][i][0] = dp[i][i][1] = num[i];
        }
        for(int len = 2; len <= n; ++len){
            for(int l = 1; l+len-1 <= 2*n; ++l){
                int r = l+len-1;
                for(int mid = l; mid < r; ++mid){ //運算符的位置
                    if(opt[mid] == 't'){ //如果是加號    
                        if(dp[l][mid][0] > -INF && dp[l][mid][0] < INF && dp[mid+1][r][0] > -INF && dp[mid+1][r][0] < INF)
                            dp[l][r][0] = min(dp[l][r][0], dp[l][mid][0]+dp[mid+1][r][0]); //最小加最小
                        if(dp[l][mid][1] > -INF && dp[l][mid][1] < INF && dp[mid+1][r][1] > -INF && dp[mid+1][r][1] < INF)
                            dp[l][r][1] = max(dp[l][r][1], dp[l][mid][1]+dp[mid+1][r][1]); //最大加最大

                    }
                    else if(opt[mid] == 'x'){ //如果是乘號
                        if(dp[l][mid][0]>-INF && dp[l][mid][0]<INF && dp[mid+1][r][0]>-INF && dp[mid+1][r][0]<INF && dp[l][mid][1]>-INF && dp[mid+1][r][1]<INF && dp[mid+1][r][1]>-INF && dp[l][mid][1]<INF){
                            dp[l][r][0] = min(dp[l][r][0], dp[l][mid][0]*dp[mid+1][r][0]); //最小乘最小
                            dp[l][r][0] = min(dp[l][r][0], dp[l][mid][1]*dp[mid+1][r][1]); //最大乘最大
                            dp[l][r][0] = min(dp[l][r][0], dp[l][mid][1]*dp[mid+1][r][0]); //最小乘最大
                            dp[l][r][0] = min(dp[l][r][0], dp[l][mid][0]*dp[mid+1][r][1]); //最大乘最小

                            dp[l][r][1] = max(dp[l][r][1], dp[l][mid][0]*dp[mid+1][r][0]); //最小乘最小
                            dp[l][r][1] = max(dp[l][r][1], dp[l][mid][1]*dp[mid+1][r][1]); //最大乘最大
                            dp[l][r][1] = max(dp[l][r][1], dp[l][mid][1]*dp[mid+1][r][0]); //最小乘最大
                            dp[l][r][1] = max(dp[l][r][1], dp[l][mid][0]*dp[mid+1][r][1]); //最大乘最小
                        }
                    }
                }
            }
        }

        ll ans = -INF;
        vector<ll> v;
        for(int l = 1; l <= n; ++l){
            if(ans < dp[l][l+n-1][1]){
                ans = dp[l][l+n-1][1];
                v.clear();
                v.push_back(l);
            }
            else if(ans == dp[l][l+n-1][1]){
                v.push_back(l);
            }
        }
        cout << ans << endl;
        int len = v.size();
        for(int i = 0; i < len; ++i)
            cout << v[i] << " ";
        cout << endl;
    }
    return 0;
}