hdu 1506

对于每一个直方块，只要能知道以它的高度往左(往右)最大能走多少，再枚举一遍所有的直方块即可。而往左(往右)最大能走多少可以用动态规划的方法实现，开数组left(right)，对于每一个方块，初始left[i](right[i])为i，然后用迭代的方法往左(右)看，即可求出。思路想出来了，代码就很简单了。

/*
 * hdu1506/win.cpp
 * Created on: 2011-10-1
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXN = 100010;

int height[MAXN], left[MAXN], right[MAXN];

void work();
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    work();
    return 0;
}

void work() {
    int N;
    long long ans, temp;
    height[0] = -1;
    while (scanf("%d", &N) == 1 && N > 0) {
        for (int i = 1; i <= N; i++) {
            scanf("%d", &height[i]);
            left[i] = i;
            right[i] = i;
        }
        height[N + 1] = -0x7fffffff;
        for (int i = 1; i <= N; i++) {
            while (height[left[i] - 1] >= height[i]) {
                left[i] = left[left[i] - 1];
            }
        }
        for (int i = N; i >= 1; i--) {
            while (height[right[i] + 1] >= height[i]) {
                right[i] = right[right[i] + 1];
            }
        }
        ans = -1;
        for (int i = 1; i <= N; i++) {
            temp = right[i] - left[i] + 1;
            temp *= height[i];
            if (temp > ans) {
                ans = temp;
            }
        }
        printf("%I64d\n", ans);
    }
}