Poj 1700 cross-river

Reference blog: http://www.cnblogs.com/steady/archive/2011/01/23/1942555.html

Http://blog.csdn.net/wuzhekai1985/article/details/6846934

The above two blogs show how to do this, which is very detailed, but not why.

I tried to prove that I felt that the strategy for this question was to send the slowest two people to the past for the shortest time. The final result of this question will be optimal.

The minimum time required to cross the river is proved. Then, every two messages are sent, and recursion is formed. But when two people are sent, is the time always the best? Why can't I send three people?

I don't know. I hope you can give me some advice.

#include<cstdio>#include<algorithm>using namespace std;int sum = 0;int Cross_river(int *input,int num){    if(num <= 2) return input[num-1];    else if(num == 3)return input[0]+input[1]+input[2];    else{        int t1=input[0]+input[1]*2+input[num-1];        int t2=input[0]*2+input[num-1]+input[num-2];        int tmp=0;        tmp = t1>t2?t2:t1;        return Cross_river(input,num-2)+tmp;    }}int main(){    int T_c,num,input[1005];    scanf("%d",&T_c);    while(T_c--)    {        scanf("%d",&num);        for(int i = 0;i < num;++i)        {            scanf("%d",&input[i]);        }        sort(input,input+num);        sum = Cross_river(input,num);        printf("%d\n",sum);    }    return 0;}