[C++]Fence Repair(貪心)
Fence Repair:
農夫約翰爲了修理柵欄,要將一塊很長的木板切割成N塊。準備切成的木塊的長度爲L1, L2,…Ln,未切割前木板的長度恰好爲切割後木板長度的總和。每次切割木板時,需要的開銷爲這塊木板的長度。例如長度爲21的木板要切成長度爲5,8,8的三塊木板。長21的木板切成長爲13和8的木板時,開銷爲21.再將長度爲13的板切成長度爲5和8的板時,開銷爲13.於是合計開銷時34.請求出按照目標要求將木板切割完最小的開銷是多少。
輸入格式:
Line 1: One integer N, the number of planks
Lines 2… N+1: Each line contains a single integer describing the length of a needed plank
輸出格式:
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
輸入:
3
8 5 8
輸出:
34
解題思路:每次選擇最小的兩塊木板相加,並將結果加入待選擇木板列中,直到只剩下一塊木板爲止。
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 20000 + 10;
int n;
ll L[maxn];
int main(){
cin>>n;
for(int i = 0; i<n; i++){
cin>>L[i];
}
sort(L, L+n);
ll sum = 0;
while(n > 1){
int m1 = 0, m2 = 1;
if(L[m1] > L[m2]){
int t = m1;
m1 = m2;
m2 = t;
}
for(int i = 2; i<n; i++){
if(L[i] < L[m1]){
m2 = m1;
m1 = i;
}
else if(L[i] < L[m2]){
m2 = i;
}
}
ll res = L[m2] + L[m1];
sum += res;
if(m1 == n-1){
int t = m1;
m1 = m2;
m2 = t;
}
L[m1] = res;
L[m2] = L[n-1];
n--;
}
cout<<sum<<endl;
return 0;
}