POJ 1179 Polygon

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题目大意：给一个多边形，每个顶点有一个值，每个边编号从1到N，边的属性是加或者乘。首先先拆掉一条边，剩下的如下做：选定一条边以及这条边的两个端点（两个数）用新顶点替换（新顶点即：按照这条边的属性（加或乘）算出这两个数的乘积或者和）。到最后剩一个点，也就是一个值。求这些值的最大值输出，并输出此时最先拆掉的是哪条边。

Input:

t -7 t 4 x 2 x 5

Output:

1 2

区间DP问题

注意最大值可以由两个最小值相乘得到

注意下标和求min_v和max_v时的逻辑简化

#include <iostream>#include <vector>using namespace std;#define SIZE 70#define MAX_V 0#define MIN_V 1#define INF 1 << 15int dp[SIZE][SIZE][2];char symbols[SIZE];int node_values[SIZE];int node_num;vector< int > v;int cal( int head, int tail, char symbol ){    if( symbol == 't' )        return head + tail;    else        return head * tail;}int main(){    cin >> node_num;    for( int i = 0; i < node_num; ++i ){        cin >> symbols[i] >> node_values[i];    }    for( int i = 0; i < node_num; ++i )        for( int j = 0; j < node_num; ++j )            if( i == j )                dp[i][i][MAX_V] = dp[i][i][MIN_V] = node_values[i];    for( int len = 1; len <= node_num - 1; ++len ){        for( int from = 0; from < node_num; ++from ){            int to = ( from + len ) % node_num;            int min_v = INF;            int max_v = -INF;            for( int count_ = 0; count_ < len; ++count_ ){                int cut = ( from + count_ ) % node_num;                int cut1 = ( from + count_ + 1 ) % node_num;                int temp = 0;                temp = cal( dp[from][cut][MAX_V], dp[cut1][to][MAX_V], symbols[cut1] );                max_v = max( temp, max_v );                temp = cal( dp[from][cut][MIN_V], dp[cut1][to][MIN_V], symbols[cut1] );                max_v = max( temp, max_v );                temp = cal( dp[from][cut][MIN_V], dp[cut1][to][MAX_V], symbols[cut1] );                min_v = min( temp, min_v );                temp = cal( dp[from][cut][MAX_V], dp[cut1][to][MIN_V], symbols[cut1] );                min_v = min( temp, min_v );            }            dp[from][to][MAX_V] = max_v;            dp[from][to][MIN_V] = min_v;        }    }    int temp = -INF;    for( int from = 0; from < node_num; ++from ){        int to = ( from + node_num - 1 ) % node_num;        if( dp[from][to][MAX_V] > temp ){                temp = dp[from][to][MAX_V];                v.clear();                v.push_back( from );        }        else if( dp[from][to][MAX_V] == temp )            v.push_back( from );    }    cout << temp <<endl;    for( int i = 0; i < v.size(); ++i )        cout << v[i] + 1 << " ";    cout << endl;    return 0;}

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