Commando War（UVA-11729）解题报告
一、原题
Time Limit: 1000ms
Case Time Limit: 1000ms
Memory Limit: 131072KB
64-bit integer IO format: %lld Java class name: Main
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G |
Commando War Input: Standard Input Output: Standard Output |
Waiting for orders we held in the wood, word from the front never came
By evening the sound of the gunfire was miles away
Ah softly we moved through the shadows, slip away through the trees
Crossing their lines in the mists in the fields on our hands and our knees
And all that I ever, was able to see
The fire in the air, glowing red, silhouetting the smoke on the breeze
There is a war and it doesn't look very promising for your country. Now it's time to act. You have a commando squad at your disposal and planning an ambush on an important enemy camp located nearby. You have N soldiers in your squad. In your master-plan, every single soldier has a unique responsibility and you don't want any of your soldier to know the plan for other soldiers so that everyone can focus on his task only. In order to enforce this, you brief every individual soldier about his tasks separately and just before sending him to the battlefield. You know that every single soldier needs a certain amount of time to execute his job. You also know very clearly how much time you need to brief every single soldier. Being anxious to finish the total operation as soon as possible, you need to find an order of briefing your soldiers that will minimize the time necessary for all the soldiers to complete their tasks. You may assume that, no soldier has a plan that depends on the tasks of his fellows. In other words, once a soldier begins a task, he can finish it without the necessity of pausing in between.
Input
There will be multiple test cases in the input file. Every test case starts with an integer N (1<=N<=1000), denoting the number of soldiers. Each of the following N lines describe a soldier with two integers B (1<=B<=10000) & J (1<=J<=10000). B seconds are needed to brief the soldier while completing his job needs J seconds. The end of input will be denoted by a case with N =0 . This case should not be processed.
Output
For each test case, print a line in the format, Case X: Y, where X is the case number & Y is the total number of seconds counted from the start of your first briefing till the completion of all jobs.
Sample Input Output for Sample Input
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3 2 5 3 2 2 1 3 3 3 4 4 5 5 0 |
Case 1: 8 Case 2: 15
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二、题目大意：
将军分配给士兵任务，每个过程都要先交代士兵任务，然后士兵处理，时间分别为begin，deal。输入n组数据，每组数据包含交代任务时间和处理任务时间，求处理完所有任务所需的最少时间。
三、思路：贪心：
将处理任务的时间从大到小排序，处理时间最长的最先处理。
原因：假设先交代X后交代Y（—表示交代时间，**表示处理时间）
①如果原先任务X比任务Y先结束，则交换他俩的位置不会得到更好的结果
X ——*** Y —————********
Y —————******** X ——***
②反之Y比X先结束
X ————*********
Y ———******
则交换后为了得到更佳的结果，必须有Y.begin+X.begin+X.deal<X.begin+Y.begin+Y.deal
Y ———******
X ————*********
即X.deal<Y.deal
循环条件：
if(end<start+sod[i].deal)
end=start+sod[i].deal;
Ps：如果end>=start+sod[i].deal就是以下情况，用时不会更长
—|————|——|(start)
****
************(end)
*****(sod[i].deal)
四、实现：对处理时间长短贪心，长的在前面
#include <iostream>
#include <algorithm>
using namespace std;
typedef struct
{
int begin;
int deal;
}Sod;
bool cmp(Sod s1,Sod s2)
{
return s1.deal>s2.deal?1:0;
}
int main()
{
int n,i,times;
Sod sod[1010];
times=1;
while(cin>>n&&n!=0)
{
int start=0,end=0;
for(i=0;i<n;i++)
{
cin>>sod[i].begin>>sod[i].deal;
}
sort(sod,sod+n,cmp);
for(i=0;i<n;i++)
{
start+=sod[i].begin;
if(end<start+sod[i].deal)
end=start+sod[i].deal;
}
cout<<"Case "<<times++<<": "<<end<<endl;
}
return 0;
}