UCF Local Programming Contest 2012(Practice) A.Wall Street Monopoly

题目链接：Wall Street Monopoly
这题读题都读了一半天，mmp…(英文果然差)。

看到这一题，我就立马联想到石子合并这一经典问题了，事实也确实如此，类似矩阵连乘，寻求最优分割点。
       0  i = j (无需合并)
dp[i][j] = { min(dp[i][j], dp[i][k] + dp[k + 1][j] + expense(i, k, j) i < j

无非再在求解合并时寻找最小代价即可。

#include <iostream>
#include <algorithm> 
#include <cstring>
using namespace std;
int n, m, ans = 1;
struct node {
	int x, y;
}a[1024];
long long dp[1024][1024];

long long Expen(int left, int mid, int right)
{
	long long exp;
	int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
	for(int i = left; i <= mid; ++i) {
		x1 += a[i].x; // 累加 
		y1 = max(y1, a[i].y); //求解最大
	}
	for(int i = mid + 1; i <= right; ++i) {
		x2 += a[i].x;  
		y2 = max(y2, a[i].y);
	}
	exp = min((long long)x1 * y2, (long long)x2 * y1);
	exp = min(exp, (long long)x1 * x2);
	exp = min(exp, (long long)y1 * y2);
	return exp * 100;
} 

void WorkPro()
{
	long long exp;
	for(int k = 1; k < m; ++k) {
		for(int i = 1; i + k <= m; ++i) { //i ~ i + k
			exp = Expen(i, i, i + k);
			dp[i][i + k] = dp[i][i] + dp[i + 1][i + k] + exp; // 小 --> 大 
			for(int j = i + 1; j < i + k; ++j) { // 切割点 
				dp[i][i + k] = min(dp[i][i + k], dp[i][j] + dp[j + 1][i + k] + Expen(i, j, i + k));
			}
		}
	}
}

int main()
{
	cin >> n;
	while(n--) {
		cin >> m;
		for(int i = 0; i < 1024; ++i)
			for(int j = 0; j < 1024; ++j)
				dp[i][j] = 0; // Clear	
		for(int i = 1; i <= m; ++i) cin >> a[i].x >> a[i].y;
		WorkPro();
		cout << "The minimum cost for lot #" << ans << " is $" << dp[1][m] << "." << endl;
		cout << endl;
		ans++;
	}
	return 0;	
}