UVA - 11729 - Commando War

来源：互联网发布：淘宝定制怎么弄编辑：程序博客网时间：2024/05/17 13:06

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=117&page=show_problem&problem=2829

题意：

有n个部下要布置，每个部下需要有对应的交待时间及任务时间，选择适当的交待顺序，以获取最小的总花费时间。

分析：

解决问题的方法不是直接模拟，而是通过数学的分析得出规律，这才是正确的思路方向；重点在于把情况正确地分类，接着进行分析。

#include <cstdio>#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>using namespace std;// #define LOCAL_TESTconst int MAX_SIZE = 1000+1;struct Soldier{int m_brief;int m_job;};Soldier szSoldiers[MAX_SIZE];int fun_compare(const void*a, const void* b){return ((Soldier*)b)->m_job - ((Soldier*)a)->m_job;}int main(){#ifdef LOCAL_TESTfreopen("f:\\in.txt", "r", stdin);// freopen("f:\\out.txt", "w+", stdout);#endifint nSoldiers;int nCase = 0;// while ( scanf("%d", &nSoldiers) == 1 && nSoldiers )while ( cin >>nSoldiers ){if ( nSoldiers == 0 )break;nCase++;memset(szSoldiers, 0, MAX_SIZE * sizeof(Soldier));for ( int i=0; i<nSoldiers; i++ )cin >>szSoldiers[i].m_brief >> szSoldiers[i].m_job;qsort(szSoldiers, nSoldiers, sizeof(szSoldiers[0]), fun_compare);// one methodint totalTime = 0;int remainTime = 0;for ( int i=0; i<nSoldiers; i++ ){totalTime += szSoldiers[i].m_brief;if ( (szSoldiers[i].m_brief + szSoldiers[i].m_job) > remainTime ){remainTime = szSoldiers[i].m_job;} // end ifelse{remainTime -= szSoldiers[i].m_brief;} // end elseif ( remainTime < 0 ){cout <<"error" <<endl;} // end if} // end fortotalTime += remainTime;cout <<"Case " <<nCase <<": " <<totalTime <<'\n';// another method-----------------------------------------------------(Best)// int totalTime = 0;// int remainTime = 0;// int sum = 0;// int ans = 0;// for ( int i=0; i<nSoldiers; i++ )// {// sum += szSoldiers[i].m_brief;// ans = max(ans, sum+szSoldiers[i].m_job);// } // end for// cout <<"Case " <<nCase <<": " <<ans <<'\n';} // end whilereturn 0;}

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