NOJ1076 Machine Dog assembly cost greedy

Title Description

SED students have recently been fascinated by the manufacture of robotic dogs and have purchased a large number of required parts that can be assembled into a component that can be assembled into a large component. In the manufacture of robotic dogs, components or parts can only be assembled in 22, in the order of assembly. In a robotic dog, each part has an assembly cost, and the cost of assembling one component at a time is the sum of the cost of each part assembly. Given the cost of each part assembly (in yuan), your task is to help sed calculate how much he spends at least.

Input

The first line includes an integer N, which indicates the number of machine dog parts (1≤n≤10000)
The second behavior is n positive integers, which indicates the cost of assembly per machine-dog part (in units of yuan), separated by a space between integers.

Output

The output is only one line, that is, the minimum cost of machine-dog assembly.

Note: The end of the output section requires an extra blank line.

Sample input

10
1 2 3 4 5 6 7 8 9 0

Sample output

136

Thinking of solving problems

The abstract of the topic is that the 22 combination of n numbers can be easily solved with greed by simply knowing that the smallest two parts of the current are grouped together. O (n^2) See Code

#include <cstdio> #include <algorithm> #define LL long long#define INF 1000000010using namespace Std;const int    MAXN = 10010;int S[maxn];int main () {int n;    scanf ("%d", &n);    for (int i = 0; i < n; i + +) scanf ("%d", &s[i]);    ll sum = 0;    if (n = = 1) {printf ("%d\n", S[0]); }else {//????        N-1??        int t = n-1; while (t--) {//??            S??????????????????            int pt1,pt2;            int _min = INF;                    for (int i = 0; i < n; i + +) {if (_min > S[i]) {pt1 = i;                _min = S[i];            }} _min = INF;                    for (int i = 0; i < n; i + +) {if (_min > s[i] && i! = pt1) {pt2 = i;                _min = S[i];            }} sum + = S[pt1] + s[pt2];            S[PT1] = s[pt1] + s[pt2];        S[PT2] = INF;    } printf ("%lld\n", sum); } return 0;} 

NOJ1076 Machine Dog assembly cost greedy