Great Graphs（小贪心）
思路： 贪心 ，以边为单位看被用了几次。
本题自己的错误： 没有把题的思路想完全，只想了一部分，只想到了到终点，而没有到其它点也可以有边。
Farmer John has a farm that consists of nn pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction. Unfortunately, Farmer John lost the map of the farm. All he remembers is an array dd , where d_id i is the smallest amount of time it took the cows to reach the ii -th pasture from pasture 11 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory. 输入格式 The first line contains one integer tt ( 1 \leq t \leq 10^41≤t≤10 4 ) — the number of test cases. Then tt cases follow. The first line of each test case contains a single integer nn ( 1 \leq n \leq 10^51≤n≤10 5 ) — the number of pastures. The second line of each test case contains nn space separated integers d_1, d_2, \ldots, d_nd 1 ,d 2 ,…,d n ( 0 \leq d_i \leq 10^90≤d i ≤10 9 ) — the array dd . It is guaranteed that d_1 = 0d 1 =0 . It is guaranteed that the sum of nn over all test cases does not exceed 10^510 5 . 输出格式 For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory. 题意翻译 给你一个长度为 nn 的序列 \{d\}{d}，你需要构造一个有向带权图，使得点 11 到点 ii 的最短路长度为 d_id i ，同时使得所有边的边权之和尽可能地小。 注意：图中不能出现负环和重边。 n\le 10^5,0\le d_i\le 10^9n≤10 5 ,0≤d i ≤10 9 输入输出样例 输入 #1复制 3 3 0 2 3 2 0 1000000000 1 0 输出 #1复制 -3 0 0 说明/提示 In the first test case, you can add roads from pasture 11 to pasture 22 with a time of 22 , from pasture 22 to pasture 33 with a time of 11 , from pasture 33 to pasture 11 with a time of -3−3 , from pasture 33 to pasture 22 with a time of -1−1 , from pasture 22 to pasture 11 with a time of -2−2 . The total cost is 2 + 1 + -3 + -1 + -2 = -32+1+−3+−1+−2=−3 .In the second test case, you can add a road from pasture 11 to pasture 22 with cost 10000000001000000000 and a road from pasture 22 to pasture 11 with cost -1000000000−1000000000 . The total cost is 1000000000 + -1000000000 = 01000000000+−1000000000=0 . In the third test case, you can't add any roads. The total cost is 00 .
#include <bits/stdc++.h> using namespace std; #define ri register int #define M 100005 template <class G>void read(G &x) { x=0;int f=0;char ch=getchar(); while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} x=f?-x:x; return ; } long long p[M]; int n,m; int main() { int T; read(T); while(T--) { read(n); for(ri i=1;i<=n;i++) { read(p[i]); } sort(p+1,p+1+n); long long ans=0; for(ri i=1;i<n;i++) { ans-=1ll*(i)*(n-i)*(p[i+1]-p[i]); } ans+=p[n]; printf("%lld\n",ans); } return 0; }
