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

151 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;}

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