HDU 4109 Instrction Arrangement（拓扑排序+bfs）

叁歲伎倆 2022-06-11 05:45 118阅读 0赞

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.

Input

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.

Output

Print one integer, the minimum time the CPU needs to run.

Sample Input

5 2
    1 2 1
    3 4 1

Sample Output

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.

题解：

先说题意：

给出一堆任务，输入三个数字x，y，z,表示完成y任务之前要完成x任务，间隔为z，每次处理要时间1，问处理完全部最少要多少时间

然后一看就是一个拓扑排序。。。然后就用dfs。。爆炸了，然后就用bfs写，WA。。后来调了半天没发现错，加个while（scanf（。。。）！=EOF）就ac了。。原来题目说了是多组数据

拓扑排序就是每次找入度为0的点，然后遍历连通的数，遍历到通往的数那个数的入度就--，然后一轮过后再找度数为0的数。。一直到结束，以前写拓扑排序可以用dfs的，这题数据有点大，只好用bfs了，这题还有点dp的意味，后来写完又觉得像求最短路径里的spfa算法的翻版---最长路径，详解看代码

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<deque>
    using namespace std;
    int p[1005][1005];//存图，两点间距离
    int a[1005];//存入度
    int num[1005];//存点通向多少边
    int to[1005][1005];//存i点通向的第j个点是哪个
    int maxx;
    int add[1005];//存最长路径
    queue<int>q;
    int main()
    {
        int i,j,k,n,m,x,y,z,tot,time;
        int now;
        while(scanf("%d%d",&n,&m)!=EOF)//不加会WA
        {
            for(i=0;i<n;i++)
            {
                a[i]=0;
                num[i]=0;
                add[i]=0;//初始化
            }
            for(i=0;i<m;i++)
            {
                scanf("%d%d%d",&x,&y,&z);//存各种信息
                p[x][y]=z;
                a[y]++;
                to[x][num[x]]=y;
                num[x]++;
            }
            tot=0;
            time=0;
            for(i=0;i<n;i++)
            {
                if(a[i]==0)//找入度为0的点
                {
                    add[i]=1;
                    q.push(i);
                }
            }
            while(!q.empty())
            {
                now=q.front();
                for(i=0;i<num[now];i++)//遍历该点通往的点
                {
                    if(add[now]+p[now][to[now][i]]>add[to[now][i]])//修改最存路径条件
                    {
                        add[to[now][i]]=add[now]+p[now][to[now][i]];
                    }
                    a[to[now][i]]--;
                    if(a[to[now][i]]==0)//入度为0了就加入
                    {
                        q.push(to[now][i]);
                    }
                }
                q.pop();
            }
            for(i=0;i<n;i++)//遍历一遍求最长
            {
                if(time<add[i])
                    time=add[i];
            }
            printf("%d\n",time);
        }
        return 0;
    }