zoj 2626 Polygon Game

    这题Dp方程很容易想到，但是居然WA了一天，后来一个博客，才发现负数对最大值的影响。恍然大悟呀……

/* * zoj_2626.cpp */#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;#define N52#define hpmax(a,b)(a)>(b)?(a):(b)#define hpmin(a,b)(a)<(b)?(a):(b)#define MIN-1e50#define MAX1e50/** * [0]: max * [1]: min, maybe negative */long long dp[N][N][2], ans;intver[N];char ope[N];int n;inline long long calculate( long long a, char oper, long long b ){if ( '+' == oper )return ( a+b );return ( a*b );}int main(){int i, j, k, len, kn, jn;long long tmp;while ( scanf("%d", &n ) != EOF && n ) {for ( i = 0; i < n; ++ i ) {scanf(" %c%d", &ope[i], &ver[i] ); }for ( i = 0; i < n; ++ i )dp[i][i][0] = dp[i][i][1] = ver[i];ans = MIN;for ( len = 1; len < n; ++ len ) {//lenght = len+1for ( i = 0; i < n; ++ i ) {j = i + len;jn = j%n;dp[i][jn][0] = MIN;dp[i][jn][1] = MAX;for ( k = i; k < j; ++ k ) {kn = k%n;tmp = calculate( dp[i][kn][0], ope[(k+1)%n], dp[(k+1)%n][jn][0] );dp[i][jn][0] = hpmax( dp[i][jn][0], tmp );dp[i][jn][1] = hpmin( dp[i][jn][1], tmp );tmp = calculate( dp[i][kn][0], ope[(k+1)%n], dp[(k+1)%n][jn][1] );dp[i][jn][0] = hpmax( dp[i][jn][0], tmp );dp[i][jn][1] = hpmin( dp[i][jn][1], tmp );tmp = calculate( dp[i][kn][1], ope[(k+1)%n], dp[(k+1)%n][jn][0] );dp[i][jn][0] = hpmax( dp[i][jn][0], tmp );dp[i][jn][1] = hpmin( dp[i][jn][1], tmp );tmp = calculate( dp[i][kn][1], ope[(k+1)%n], dp[(k+1)%n][jn][1] );dp[i][jn][0] = hpmax( dp[i][jn][0], tmp );dp[i][jn][1] = hpmin( dp[i][jn][1], tmp );}if ( n-1 == len )ans = hpmax( ans, dp[i][jn][0] );}}printf("%lld\n", ans );}return 0;}