图论--差分约束--HDU\HDOJ 4109 Instrction Arrangement

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.

题意：一台电脑需要执行N条指令（0到N-1），每条指令都要花费一单位时间，可以同时执行无限条指令。有M个约束条件（X,Y,Z），表示指令Y必须在指令X执行后过Z单位时间才能执行。问执行完所有的指令需要的最短时间。

思路：显然就是差分约束嘛，设Si为指令i的开始时间，对每条约束可以得到不等式 Sy >= Sx + Z。

这道题目建的图不一定是连通的，采用初始时将所有结点加入队列的方法代替超级源点，将图变成虚连通的。

求最短时间所有跑最长路，因为图中不含负边，所以可以直接将初始的距离dis置为1，不用置为-inf。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define INF 1e9
using namespace std;
const int maxn =1000+10;
const int maxm =20000+10;
 
struct Edge
{
    int from,to,dist;
    Edge(){}
    Edge(int f,int t,int d):from(f),to(t),dist(d){}
};
 
struct BellmanFord
{
    int n,m;
    int head[maxn],next[maxm];
    Edge edges[maxm];
    int d[maxn];
    bool inq[maxn];
 
    void init(int n)
    {
        this->n=n;
        m=0;
        memset(head,-1,sizeof(head));
    }
 
    void AddEdge(int from,int to,int dist)
    {
        edges[m]=Edge(from,to,dist);
        next[m]=head[from];
        head[from]=m++;
    }
 
    void bellmanford()
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;i++) d[i]= i==0?0:INF;
        queue<int> Q;
        Q.push(0);
 
        while(!Q.empty())
        {
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for(int i=head[u];i!=-1;i=next[i])
            {
                Edge &e=edges[i];
                if(d[e.to] > d[u]+e.dist)
                {
                    d[e.to] = d[u]+e.dist;
                    if(!inq[e.to])
                    {
                        inq[e.to]=true;
                        Q.push(e.to);
                    }
                }
            }
        }
    }
}BF;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        BF.init(n+1);
        while(m--)
        {
            int u,v,d;
            scanf("%d%d%d",&u,&v,&d);
            ++u,++v;
            BF.AddEdge(v,u,-d);
        }
        for(int i=1;i<=n;i++)
            BF.AddEdge(0,i,0);
        BF.bellmanford();
        int max_v=-1,min_v=INF;
        for(int i=1;i<=n;i++)
        {
            max_v=max(max_v,BF.d[i]);
            min_v=min(min_v,BF.d[i]);
        }
        printf("%d\n",max_v-min_v+1);
    }
    return 0;
}