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  • 【UOJ #244】【UER #7】短路

    http://uoj.ac/contest/35/problem/244
    对其他人来说好简单的一道题,我当时却不会做TWT
    注定滚粗啊
    题解很好的~

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    const int N = 100003;
    ll in() {
    	ll k = 0, fh = 1; char c= getchar();
    	for (; c < '0' || c > '9'; c = getchar())
    		if (c == '-') fh = -1;
    	for (; c >= '0' && c <= '9'; c = getchar())
    		k = k * 10 + c - 48;
    	return k * fh;
    }
    
    int n;
    ll a[N], f[N], ans, now;
    
    int main() {
    	scanf("%d", &n);
    	for (int i = 0; i <= n; ++i) a[i] = in();
    	f[n] = a[n];
    	now = a[n];
    	ans = a[n] * (4 * n + 1);
    	for (int i = n - 1; i >= 0; --i) {
    		f[i] = f[i + 1] + now + a[i];
    		if (a[i] < now) {
    			now = a[i];
    			ans = min(ans, f[i] * 2 + now * (4 * i - 1));
    		}
    	}
    	printf("%lld
    ", ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/abclzr/p/5972029.html
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