2015
04-16

# Instrction Arrangement

Ali has taken the Computer Organization and Architecture course this term. He learned that there may be dependence between instructions, like WAR (write after read), WAW, RAW.
If the distance between two instructions is less than the Safe Distance, it will result in hazard, which may cause wrong result. So we need to design special circuit to eliminate hazard. However the most simple way to solve this problem is to add bubbles (useless operation), which means wasting time to ensure that the distance between two instructions is not smaller than the Safe Distance.
The definition of the distance between two instructions is the difference between their beginning times.
Now we have many instructions, and we know the dependent relations and Safe Distances between instructions. We also have a very strong CPU with infinite number of cores, so you can run as many instructions as you want simultaneity, and the CPU is so fast that it just cost 1ns to finish any instruction.
Your job is to rearrange the instructions so that the CPU can finish all the instructions using minimum time.

The input consists several testcases.
The first line has two integers N, M (N <= 1000, M <= 10000), means that there are N instructions and M dependent relations.
The following M lines, each contains three integers X, Y , Z, means the Safe Distance between X and Y is Z, and Y should run after X. The instructions are numbered from 0 to N – 1.

The input consists several testcases.
The first line has two integers N, M (N <= 1000, M <= 10000), means that there are N instructions and M dependent relations.
The following M lines, each contains three integers X, Y , Z, means the Safe Distance between X and Y is Z, and Y should run after X. The instructions are numbered from 0 to N – 1.

5 2
1 2 1
3 4 1

2
Hint
In the 1st ns, instruction 0, 1 and 3 are executed;
In the 2nd ns, instruction 2 and 4 are executed.
So the answer should be 2.


#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define val 1001
{
int v;
int cost;
};
struct node
{
}vode[val];
int into[val],ee[val],n,m;//into表示入度，ee表示从最早完成该任务的时间，ee初始化为1
void topsort();//拓扑排序，边拓扑边求最长路径
int main()
{
int i,j,ans;
while(scanf("%d %d",&n,&m)!=EOF)
{
memset(into,0,sizeof(into));
for(i=0;i<n;i++)
ee[i]=1;
ans=0;
topsort();
for(i=0;i<n;i++)
if(ee[i]>ans) ans=ee[i];
printf("%d\n",ans);
}
return 0;
}
{
int i,j,a,b,c;
for(i=0;i<n;i++)
vode[i].next=NULL;
for(i=0;i<m;i++)
{
scanf("%d %d %d",&a,&b,&c);
into[b]++;
s->v=b;
s->cost=c;
s->next=vode[a].next;
vode[a].next=s;
}
}
void topsort()
{
int i,j,v;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
if(into[j]==0)
{
v=j;
break;
}
into[v]=-1;
p=vode[v].next;
while(p!=NULL)
{
if(ee[p->v]<ee[v]+p->cost)//如果权值和更大，则替换
ee[p->v]=ee[v]+p->cost;
into[p->v]--;
p=p->next;
}
}
}


1. 额~阅片无数如小小叨，也只能想到他很久以前的一部作品——《放羊的星星》。李威在剧里饰演痴心悲剧男二，男一是林志颖演的哦~

2. 我没看懂题目
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
第二个是7 0 6 -1 1 -6 7输出是14 7 7
不知道题目例子是怎么得出来的