Crossing River
A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.
Output
For each test case, print a line containing the total number of seconds required for all the N people to cross the river.
Sample Input
1
4
1 2 5 10
Sample Output
17
Source
题目类型：贪心
算法分析：如果n比较小则会比较容易得到答案，如果n比较大则对于每一次递推都需要判断两种方案中哪一个更优，其中一种方案是最快的来回运送最慢的，另一种方案是最快的和次快的一起过河，然后最快的回来。然后最慢的和次慢的一起过河，次快的回来
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 1000000000 + 7; const int maxn = 1000 + 666; int ans[maxn], dp[maxn]; int main() { // ifstream cin ("aaa.txt"); int t; cin >> t; while (t--) { int n; cin >> n; for (int i = 0; i < n;i++) cin >> ans[i]; sort (ans, ans + n); dp[0] = ans[0], dp[1] = ans[1]; int minvala, minvalb; for (int i = 2; i < n; i++) { minvala = dp[i-1] + ans[0] + ans[i]; minvalb = dp[i-2] + ans[0] + 2 * ans[1] + ans[i]; dp[i] = min (minvala, minvalb); } cout << dp[n-1] << endl; } return 0; } |
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