ZOJ-2422

题目看了半天也看不懂，反正和1985那个题差不多就是了，都是求并排不等高矩形的最大矩形面积，唯一区别就是那个题宽度固定为1，本题宽度不固定，代码都差不多，加上宽度的处理就行了，把那题代码拿过来稍微改改就A了。。

#include<stdio.h>

int main()
{
	int n, w[50000], h[50000], left[50000], right[50000], lw[50000], rw[50000];
	while (scanf("%d", &n), n != -1)
	{
		int i;
		for (i = 0; i < n; i++)
		{
			scanf("%d %d", &w[i], &h[i]);
			left[i] = i;
			right[i] = i;
			lw[i] = rw[i] = w[i];
		}
		int temp;
		for (i = 0; i < n; i++)
		{
			temp = i;
			while (temp - 1 >= 0 && h[temp - 1] >= h[i])
			{
				lw[i] += lw[temp - 1];
				temp = left[temp - 1];
			}
			left[i] = temp;
		}
		for (i = n - 1; i >= 0; i--)
		{
			temp = i;
			while (temp + 1 < n && h[temp + 1] >= h[i])
			{
				rw[i] += rw[temp + 1];
				temp = right[temp + 1];
			}
			right[i] = temp;
		}
		int max = 0, res;
		for (i = 0; i < n; i++)
		{
			res = (lw[i] + rw[i] - w[i]) * h[i];
			if (res > max)
				max = res;
		}
		printf("%d\n", max);
	}
	return 0;
}