HDU 3572 Task Schedule（拆点+最大流dinic）

Posted 2020-08-22 Blackops

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7753    Accepted Submission(s): 2381

Problem Description

Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days. 
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.

 

 

Input

On the first line comes an integer T(T<=20), indicating the number of test cases.

You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.

 

 

Output

For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.

Print a blank line after each test case.

 

 

Sample Input

2

4 3

1 3 5

1 1 4

2 3 7

3 5 9

 

2 2

2 1 3

1 2 2

 

 

Sample Output

Case 1: Yes

Case 2: Yes

 

 

题目链接：HDU 3572

拆点的最大流判断是否满流的题目，点怎么拆呢？从源点S连向每一个任务i一条容量为p的边，说明每一个任务一开始要p个流量流入，然后每一个任务i向时间点[s,e]连一条容量为1的边，说明一个任务只能在一个机器上被工作，即不能既在机器A上加工又在机器B上加工，然后每一个时间点向T连一条容量为m个边，说明一个时间点只能最多同时有m个机器在工作。最后你就是要判断从S流出的$n*p$个流量能否全部流入T中就好了

空间复杂度大概是$（500+500^2+500）*2$条边，$500+500$个点，原本只会最辣鸡的FF想低空卡过这题，然而被无限TLE教做人，查查题解又膜膜dinic，发现dinic也容易理解，分层的意义就是减少没有用的搜索，因为增广一定是从最小距离距离近的到最小距离远的，那么那些d[v]!=d[u]+1的点就可以被忽略掉了

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1010;
const int M=251000+7;
struct edge
{
    int to,nxt;
    int cap;
};
edge E[M<<1];
int head[N],tot,d[N];

void add(int s,int t,int cap)
{
    E[tot].to=t;
    E[tot].cap=cap;
    E[tot].nxt=head[s];
    head[s]=tot++;

    E[tot].to=s;
    E[tot].cap=0;
    E[tot].nxt=head[t];
    head[t]=tot++;
}
void init()
{
    CLR(head,-1);
    tot=0;
}
int bfs(int s,int t)
{
    CLR(d,-1);
    d[s]=0;
    queue<int>Q;
    Q.push(s);
    while (!Q.empty())
    {
        int now=Q.front();
        Q.pop();
        for (int i=head[now]; ~i; i=E[i].nxt)
        {
            int v=E[i].to;
            if(d[v]==-1&&E[i].cap>0)
            {
                d[v]=d[now]+1;
                if(v==t)
                    return 1;
                Q.push(v);
            }
        }
    }
    return d[t]!=-1;
}
int dfs(int s,int t,int f)
{
    if(s==t||!f)
        return f;
    int r=0;
    for (int i=head[s]; ~i; i=E[i].nxt)
    {
        int v=E[i].to;
        if(d[v]==d[s]+1&&E[i].cap)
        {
            int d=dfs(v,t,min(f,E[i].cap));
            if(d>0)
            {
                E[i].cap-=d;
                E[i^1].cap+=d;
                r+=d;
                f-=d;
                if(!f)
                    break;
            }
        }
    }
    if(!r)
        d[s]=INF;
    return r;
}
int dinic(int s,int t)
{
    int r=0;
    while (bfs(s,t))
        r+=dfs(s,t,INF);
    return r;
}
int main(void)
{
    int tcase,p,s,e,i,j,n,m;
    scanf("%d",&tcase);
    for (int q=1; q<=tcase; ++q)
    {
        init();
        scanf("%d%d",&n,&m);
        int S=0;
        int tl=INF,tr=-INF;
        int sump=0;
        for (i=1; i<=n; ++i)
        {
            scanf("%d%d%d",&p,&s,&e);
            add(S,i,p);
            sump+=p;

            if(s<tl)
                tl=s;
            if(e>tr)
                tr=e;

            for (j=s; j<=e; ++j)
                add(i,n+j,1);
        }
        int T=n+tr+1;
        for (i=tl; i<=tr; ++i)
            add(n+i,T,m);
        printf("Case %d: %s\n\n",q,dinic(S,T)==sump?"Yes":"No");
    }
    return 0;
}