题目描述
Afandi is herding N sheep across the expanses of grassland when he finds himself blocked by a river. A single raft is available for transportation.
Afandi knows that he must ride on the raft for all crossings, but adding sheep to the raft makes it traverse the river more slowly.
When Afandi is on the raft alone, it can cross the river in M minutes When the i sheep are added, it takes Mi minutes longer to cross the river than with i-1 sheep (i.e., total M+M1 minutes with one sheep, M+M1+M2 with two, etc.).
Determine the minimum time it takes for Afandi to get all of the sheep across the river (including time returning to get more sheep).
输入
On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 5 Each case contains:
* Line 1: one space-separated integers: N and M (1 ≤ N ≤ 1000 , 1≤ M ≤ 500).
* Lines 2..N+1: Line i+1 contains a single integer: Mi (1 ≤ Mi ≤ 1000)
输出
For each test case, output a line with the minimum time it takes for Afandi to get all of the sheep across the river.
样例输入
2
2 10
3
5
5 10
3
4
6
100
1
样例输出
18 50
要认真读题才能看懂题意：
题中说的是如果第一次带两只羊，时间是m+m1+m2;第二次带一只羊用时m1（第一只羊的m）；第三次带2只羊用m1+m2（第一只羊和第二只羊的m）；
例如第二组数据：第一次带3+4+6+10=23；第二次3+4+10=17；共用时23+17+10=50；
用一个前缀和数组time，time[i]表示单独运送i只羊所花费的时间。
dp[i]表示一个人和i只羊过河所花费的最短时间，则开始时dp[i] = time[i] + M，
以后更新时，dp[i] = min（dp[i]，dp[i-j] + m + dp[j]），
j从1循环到i-1，即把i只羊分成两个阶段来运，只需求出这两个阶段的和，
然后加上人从对岸回来所用的时间，与dp[i]进行比较，取最小值。
#include<stdio.h> #include<algorithm> using namespace std; int dp[1005], time[1005]; int main() { int T, n, m, i, j; scanf("%d",&T); while(T--) { int a; scanf("%d%d",&n, &m); time[0] = 0; for(i = 1; i <= n; i++) { scanf("%d",&a); time[i] = time[i-1] + a; } for(i = 1; i <= n; i++) { dp[i] = time[i] + m; for(j = 1; j < i; j++) dp[i] = min(dp[i], dp[i-j] + dp[j] + m); } printf("%d ",dp[n]); } return 0; } //time[i]表示一次运送i只羊所花费的时间 //dp[i]表示人和i只羊一起过河所花费的最短时间