CF5E Bindian Signalizing

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  这题的意思是给一个环状的山的高度，如果两座山互相能看见便符合，求符合的对数有多少。这里l[i]为i左边第一个比i大的山的位置，r[i]为i右边第一个比i大的山的位置，v[i]为i右边在遇到比i高的山之前与i相同的山的个数.

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#define maxn 1000005#define mem(a, b) memset(a, b, sizeof(a))using namespace std;__int64 num1[maxn], num2[maxn], n, ans;int l[maxn], r[maxn], v[maxn];int main(int argc, char *argv[]){int flag, maxs, w;while(~scanf("%I64d", &n)){maxs = -1;w = 0;ans = 0;mem(v, 0);for(int i = 0;i < n;i++){scanf("%I64d", &num1[i]);if(maxs < num1[i]){maxs = num1[i];flag = i;}}for(int i = flag;i < n;i++)num2[w++] = num1[i];for(int i = 0;i < flag;i++)num2[w++] = num1[i];num2[w] = maxs;for(int i = 1;i < n;i++){l[i] = i - 1;while(l[i]&&num2[i] >= num2[l[i]])l[i] = l[l[i]];}for(int i = (n - 1);i >= 0;i--){r[i] = i + 1;while(r[i] != n&&num2[i] > num2[r[i]])r[i] = r[r[i]];if(num2[i] == num2[r[i]]&&r[i] != n){v[i] = v[r[i]] + 1;r[i] = r[r[i]];}}for(int i = 1;i < n;i++){ans += v[i];ans += 2;if(!l[i]&&r[i] == n)ans--;}printf("%I64d\n", ans);}return 0;}

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