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2014湘潭邀请赛 C题 湘大OJ 1205 Range (单调栈)
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)
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