Crossing River（贪心 博弈）

Crossing River
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Status
Description
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

题意：多人过河，没人速度都不同，小船一次最多载两个人，速度取两人中较慢的一个。
首行给出T，代表有T组数据，每组数据第一行给出N，代表接着一行有N个人的过河时间，输出最小过河时间。

思路：两种过河方案
一：最快次快去，最快回，最慢次慢去，次快回，最慢次慢过河了。
二：最快最慢去，最快回，最快次慢去，最快回，最慢次慢过河了。
每轮可以送过去两个人，每次取耗时较少的方案，直至剩余人数小于4，再特判剩下的几个人即可。

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>#include<queue>using namespace std;//两种策略const int maxn=1005;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int N;        scanf("%d",&N);        int num[maxn];        for(int i=0; i<N; i++)            scanf("%d",&num[i]);        sort(num,num+N);        int result=0;//记录结果        while(N>3)//剩余数量小于4，构不成一个循环        {            //一：最快次快去，最快回，最慢次慢去，次快回            //二：最快最慢去，最快回，最快次慢去，最快回            result+=min(num[1]+num[0]+num[N-1]+num[1],num[N-1]+num[0]+num[N-2]+num[0]);            N-=2;        }        if(N==1)//还剩一个            result+=num[0];        else if(N==2)//还剩两个            result+=num[1];        else//还剩三个            result+=num[0]+num[1]+num[2];        printf("%d\n",result);    }    return 0;}

和NYOJ过河问题一模一样http://blog.csdn.net/qq_32680617/article/details/50676472