2014湘潭邀请赛 C题 湘大OJ 1205 Range （单调栈）

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Problem Description

For an array, the range function is defined below: Range(A)=Max(A)-Min(A)+1; For example, suppose A={1,2,3,4,5}, then Range(A)=5-1+1=5. Now, given an array A(length≤100000), you are going to calcalute the sum of all subarray‘s range. i.e sigma(i,j){Range(A[i,j])}.

Input

First line contain an integer T, there are T(1≤T≤100) cases. For each case T. The length N(1≤N≤100000), and N integers A[i](1≤A[i]≤10⁹).

Output

Output case number first, then the answer.

Sample Input

1
5
1 2 3 4 5

Sample Output

Case 1: 35

实际上就是求出以第i个数为最大值得区间个数 和 以第i个数为最小值得区间个数 ，可以利用单调栈的思想，和poj2559几乎是一个题。都是求出以第i个为最值，往左往右能扩展出的范围。

#include <iostream>
#include <cstdio>
using namespace std;
typedef __int64 LL;
const int maxn = 100000+10;
int t, n;
LL l[maxn], r[maxn], a[maxn];
LL solve()
{
    LL ans = (LL)(n+1)*n/2;
    l[1] = 1;
    for(int i = 2; i <= n; i++) {
        int tmp = i;
        while(tmp > 1 && a[tmp-1] >= a[i]) tmp = l[tmp-1];
        l[i] = tmp;
    }
    r[n] = n;
    for(int i = n-1; i > 0; i--) {
        int tmp = i;
        while(tmp < n && a[tmp+1] > a[i]) tmp = r[tmp+1];
        r[i] = tmp;
    }
    for(int i = 1; i <= n; i++) ans -= a[i] * (LL)(i - l[i] + 1) * (LL)(r[i] - i + 1);
    l[1] = 1;
    for(int i = 2; i <= n; i++) {
        int tmp = i;
        while(tmp > 1 && a[tmp-1] <= a[i]) tmp = l[tmp-1];
        l[i] = tmp;
    }
    r[n] = n;
    for(int i = n-1; i > 0; i--) {
        int tmp = i;
        while(tmp < n && a[tmp+1] < a[i]) tmp = r[tmp+1];
        r[i] = tmp;
    }
    for(int i = 1; i <= n; i++) ans += a[i] * (LL)(i - l[i] + 1) * (LL)(r[i] - i + 1);

    return ans;
}
int main()
{
    cin >> t;
    for(int ca = 1; ca <= t; ca++) {
        cin >> n;
        for(int i = 1; i < n+1; i++) scanf("%I64d", &a[i]);
        printf("Case %d: %I64d\n", ca, solve());
    }
    return 0;
}

2014湘潭邀请赛 C题 湘大OJ 1205 Range （单调栈）

标签：des   style   blog   os   io   ar   for   2014   div   

原文地址：http://blog.csdn.net/u013923947/article/details/39034817