Crossing River poj1700 Greedy

Title Description: n Individuals across the river, there is only one boat, up to two people rowing, each rowing speed is different, the speed of the boat speed of the slowest person. Input T is the number of case, each case input n is the number of people, the next line of input is the time each person crosses the river, are not the same. The time required to output n people all across the river

Algorithm thought: Adopt the Greedy method. There are two ways of rowing, one is the fastest + slowest, the fastest back, the fastest + times slow, the fastest back, cycle down, the second is the fastest + times faster, faster back, the slowest + times slow, the fastest back, cycle down. So when the rest of the population n>3, compare the first and second time, using a short-time method to cross the river. n=1,2,3 the time to discuss it alone.

#include <iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespacestd;intpt[1010];//Store time of peopleintMain () {intT; scanf ("%d",&T);  while(t--){        intN; scanf ("%d",&N); Memset (PT,0,sizeof(PT));  for(intI=0; i<n;i++) {scanf ("%d", pt+i); } sort (Pt,pt+N); intt=0;  while(n>3){            if(2*pt[1]+pt[0]>2*pt[0]+pt[n-2]) {T+=2*pt[0]+pt[n-1]+pt[n-2]; }            Else{T+=2*pt[1]+pt[n-1]+pt[0]; } N-=2; }        if(n==3) {T+=pt[2]+pt[0]+pt[1]; }        Else if(n==2) {T+=pt[1]; }        Else{T+=pt[0]; } printf ("%d\n", T); }    return 0;}

Crossing River poj1700 Greedy