poj3253 Fence Repair

/* 解析见挑战P48-49  法一：递归--复杂度(O(n^2))*/#include <iostream>typedef long long ll;const int MAX_N = 2e4 + 10;int N, L[MAX_N];using namespace std;void show(){for (int i = 0; i < N; i++)cout << L[i] << " ";cout << endl;}void solve(){ll ans = 0;// 直到计算模板为1块为止while ( N > 1 ){// 求出最短的板的下标mii1，和次短的板的下标mii2int mii1 = 0, mii2 = 1; if ( L[mii1] > L[mii2] ) swap(mii1, mii2);for ( int i = 2; i < N; i++ ){if ( L[i] < L[mii1] ){mii2 = mii1;mii1 = i;}else if ( L[i] < L[mii2] ){mii2 = i;}}//show();// 将两块板拼合int t = L[mii1] + L[mii2];ans += t;//cout << L[mii1] << " " << L[mii2] << endl;//cout << "ans = " << ans << endl;if (mii1 == N - 1) swap(mii1, mii2);L[mii1] = t;L[mii2] = L[N - 1];//cout << "after changed: " << L[mii1] << " " << L[mii2] << endl;//cout << "mii1 = " << mii1 << " mii2 = " << mii2 << endl;//show();N--; }cout << ans << endl; }int main(){cin >> N;for ( int i = 0; i < N; i++ ){cin >> L[i];}solve( );return 0;}

//法二：见挑战 P77 //改进：使用优先队列 复杂度O(nlogn)#include <iostream>#include <queue>typedef long long ll;const int MAX_N = 2e4 + 10;int N, L[MAX_N];using namespace std;typedef long long ll;typedef priority_queue<int, vector<int>, greater<int> > QUEUE;//void test(QUEUE a)//{//while (a.size())//{//cout << a.top() << endl;//a.pop();//}//}void solve(){ll ans = 0;// 声明一个从小到大取出数值的优先队列QUEUE que;for (int i = 0; i < N; i++)que.push(L[i]);//test(que);// 循环到只剩下一块木板为止while (que.size() > 1){// 取出最短的木板和次短的木板int l1, l2;l1 = que.top();que.pop();l2 = que.top();que.pop();// 把两块木板合并ans += l1 + l2;que.push(l1 + l2); //test(que);} cout << ans << endl;}int main(){cin >> N;for ( int i = 0; i < N; i++ ){cin >> L[i];}solve( );return 0;}