[贪心]ZOJ1012、1076

ZOJ 1012:

http://acm.zju.edu.cn/onlinejudge/showRuns.do?contestId=1

题目不是很难，读懂题之后觉得按开始时间排序，相同时间的按酬劳排序，依次处理即可。

其实我不太明白也不能证明，为什么这种活动安排或者资源配置的题，按贪心都能解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;

struct job
{
int cpu,mem;
int stime,etime;
int award,extra,punlish;
bool operator<(const job& other)const
{
if ( stime==other.stime )
return award>other.award;
else
return stime<other.stime;
}
};

int fTime;
int m,n,l;
int ans;
int nCase;

job jobs[10005];
int finish[10005];

void solve()
{
ans=0;
int i=0;
int curTime=jobs[0].stime;
int first=curTime;
int ccpu=m,cmem=n;
for(;curTime<fTime;curTime++)
{
ccpu=m;cmem=n;
for (i=0;i<l;i++)
{
if(jobs[i].stime>curTime)
break;
if (!finish[i]&&ccpu>=jobs[i].cpu&&cmem>=jobs[i].mem)
{
finish[i]=1;
ccpu-=jobs[i].cpu;
cmem-=jobs[i].mem;
ans+=jobs[i].award;
if (curTime+1<=jobs[i].etime)
ans+=jobs[i].extra*(jobs[i].etime-curTime-1);
else
ans-=jobs[i].punlish*(-jobs[i].etime+curTime+1);
}
}
}
for(i=0;i<l;i++)
{
if(jobs[i].stime>fTime)
break;
if (!finish[i] && jobs[i].etime<fTime)
{
ans-=jobs[i].punlish*(fTime-jobs[i].etime);
}
}
printf("Case %d: %d\n\n",nCase,ans);
}
void init()
{
memset(finish,0,sizeof(finish));
scanf("%d%d%d",&m,&n,&l);
for(int i=0;i<l;i++)
{
job& j=jobs[i];
scanf("%d%d%d%d%d%d%d",&j.cpu,&j.mem,&j.stime,&j.etime,&j.award,&j.extra,&j.punlish);
}
sort(&jobs[0],&jobs[l]);
}
int main()
{
nCase=0;
while(scanf("%d",&fTime)&&fTime!=0)
{
nCase++;
init();
solve();
}
}

ZOJ1076:

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=76

题目其实就是活动安排，比较简单，当然也可以看做求最长递增子序列。

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
using namespace std;

struct gene
{
int start,end;
int idx;
bool operator<(const gene& other)const
{
//	if ( end==other.end )
//	return start<other.start;
return end<other.end;
}
};

gene gens[1005];
int use[1005];
int n;

int main()
{
while(scanf("%d",&n)&&n)
{
memset(use,0,sizeof(use));
int i;
for(i=0;i<n;i++)
{
scanf("%d%d",&gens[i].start,&gens[i].end);
gens[i].idx=i+1;
}
sort(gens,gens+n);

use[0]=1;
int time=gens[0].end;
for(i=1;i<n;i++)
{
if(gens[i].start>time)
{
use[i]=1;
time=gens[i].end;
}
}

printf("%d",gens[0].idx);
for(i=1;i<n;i++)
if(use[i])
printf(" %d",gens[i].idx);
printf("\n");
}
}