hdu 4109 Instrction Arrangement 拓扑排序/觉得更像BFS

Instrction Arrangement

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1385 Accepted Submission(s): 576



Problem Description

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

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.

Source

2011 Alibaba-Cup Campus Contest



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#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)
#define mp    make_pair
#define _f     first
#define _s     second

using namespace std;
const int INF =0x3f3f3f3f;
const int maxn=  1000+8   ;
const int maxm=10000+11    ;
//const int INF=    ;
typedef long long ll;
const ll inf =1000000000000000;//1e15;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
//by yskysker123
int n,m;
int e_max;
int u[maxm],v[maxm],w[maxm];
int nex[maxm];
int fir[maxn];
int d[maxn];
int dis[maxn];

int q[maxn];
void init()
{
    memset(fir,-1,sizeof fir);
    e_max=0;
}

inline void add_edge(int s,int t,int w_)
{
    int e=e_max++;
    u[e]=s;
    v[e]=t;
    w[e]=w_;
    nex[e]=fir[s];
    fir[s]=e;

}

void top()
{
    memset(d,0,sizeof d);
    for(int e=0;e<e_max;e++)
    {
        int y=v[e];
        d[y]++;
    }
    int f=0,r=0;
    for(int i=0;i<n;i++)
    {
        dis[i]=1;
        if(d[i]) continue;
          q[r++]=i;
    }
    while( f<r)
    {
       int x=q[--r];
       int dist=dis[x];
        for(int e=fir[x];~e;e=nex[e])
        {
            int y=v[e];
            dis[y]=max(dis[y], dist+ w[e]);
            if(--d[y ]==0)
            {

                q[r++]=y;
            }

        }
    }

}
int main()
{
    int x,y,w;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&w);
            add_edge(x,y,w);
        }
        top();
        int maxi=1;
        for(int i=0;i<n;i++)
            maxi=max(dis[i],maxi);
        printf("%d\n",maxi);

    }

    return 0;
}