POJ 2796 Feel Good (单调栈)

http://poj.org/problem?id=2796

和POJ 2559一样，把计算矩形的长改为前缀和就行。

完整代码：

/*891ms,2716KB*/#include<cstdio>#include<algorithm>using namespace std;const int mx = 100005;__int64 h[mx], sum[mx];int l[mx], r[mx];int main(){//freopen("in.txt", "r", stdin);int n, i, j, ansl, ansr;__int64 ans, tmp;scanf("%d", &n);for (i = 1; i <= n; ++i){scanf("%I64d", &h[i]);sum[i] = sum[i - 1] + h[i];}for (i = 1; i <= n; ++i){j = i;while (j > 1 && h[j - 1] >= h[i]) ///压栈否j = l[j - 1]; ///出栈l[i] = j;}for (i = n; i; --i){j = i;while (j < n && h[j + 1] >= h[i]) ///压栈否j = r[j + 1]; ///出栈r[i] = j;}ans = -1;///注意数据中有0for (i = 1; i <= n; ++i){tmp = h[i] * (sum[r[i]] - sum[(l[i] - 1)]);if (tmp > ans){ans = tmp;ansl = l[i];ansr = r[i];}}printf("%I64d\n%d %d\n", ans, ansl, ansr);return 0;}