hdoj 3572 Task Schedule【建立超级源点超级汇点】

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6000    Accepted Submission(s): 1922

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

 

题意：有n个任务，m个机器，给你完成第i件任务的时间以及必须完成这件任务的时间区间（si，ei），一台机器一次只能执行一个任务，让你判断是否完成所有的任务；

 

题解：难在建图   注意： 我们是将天数看做流量

1、将每个任务i看做一个节点连接超级源点 s，容量为每个人物所需要的时间

2、将每个任务i看做节点， 连接完成这个任务所要进行的时间阶段内的所有点，容量为1  （表示这件任务的流量只能为1（即天数为1））

3、将所有时间段内的点连接到超级汇点t容量为m   （表示一天内共有m台机器可以同时工作）

 

 

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#include<stdio.h> #include<string.h> #include<queue> #include<stack> #include<algorithm> #define INF 0x7fffff #define MAX 11000 #define MAXM 1001000 using namespace std; struct node {     int from,to,cap,flow,next; }edge[MAXM]; int ans,head[MAX]; int cur[MAX]; int vis[MAX]; int dis[MAX]; int sum,n,m; int sec;//超级汇点 void init() {     ans=0;     memset(head,-1,sizeof(head)); } void add(int u,int v,int w) {     node E1={u,v,w,0,head[u]};     edge[ans]=E1;     head[u]=ans++;     node E2={v,u,0,0,head[v]};     edge[ans]=E2;     head[v]=ans++; } void getmap() {     int i,j,last=-1;     sum=sec=0;     int bt,et,time;     for(i=1;i<=n;i++)     {         scanf("%d%d%d",&time,&bt,&et);         sum+=time;         add(0,i,time);//超级源点连接第i件任务         for(j=bt;j<=et;j++)             add(i,n+j,1);//将每件任务与完成这件任务所需要的时间段内的每一天连接         last=max(last,et);     }     sec=n+last+1;     for(i=1;i<=sec;i++)         add(n+i,sec,m);//将所有的时间段内的点指向超级汇点 } int bfs(int beg,int end) {     int i;     memset(vis,0,sizeof(vis));     memset(dis,-1,sizeof(dis));     queue<int>q;     while(!q.empty())         q.pop();     vis[beg]=1;     dis[beg]=0;     q.push(beg);     while(!q.empty())     {         int u=q.front();         q.pop();         for(i=head[u];i!=-1;i=edge[i].next)//遍历所有的与u相连的边         {             node E=edge[i];             if(!vis[E.to]&&E.cap>E.flow)//如果边未被访问且流量未满继续操作             {                 dis[E.to]=dis[u]+1;//建立层次图                 vis[E.to]=1;//将当前点标记                                                                                                                            if(E.to==end)//如果当前点搜索到终点则停止搜索  返回1表示有从原点到达汇点的路径                     return 1;                 q.push(E.to);//将当前点入队             }         }     }     return 0;//返回0表示未找到从源点到汇点的路径 } int dfs(int x,int a,int end)//把找到的这条边上的所有当前流量加上a（a是这条路径中的最小残余流量） {     //int i;     if(x==end||a==0)//如果搜索到终点或者最小的残余流量为0         return a;     int flow=0,f;     for(int& i=cur[x];i!=-1;i=edge[i].next)//i从上次结束时的弧开始     {         node& E=edge[i];         if(dis[E.to]==dis[x]+1&&(f=dfs(E.to,min(a,E.cap-E.flow),end))>0)//如果         {//bfs中我们已经建立过层次图，现在如果 dis[E.to]==dis[x]+1表示是我们找到的路径         //如果dfs>0表明最小的残余流量还有，我们要一直找到最小残余流量为0             E.flow+=f;//正向边当前流量加上最小的残余流量             edge[i^1].flow-=f;//反向边             flow+=f;//总流量加上f             a-=f;//最小可增流量减去f             if(a==0)                 break;         }     }     return flow;//所有边加上最小残余流量后的值 } int Maxflow(int beg,int end) {     int flow=0;     while(bfs(beg,end))//存在最短路径     {         memcpy(cur,head,sizeof(head));//复制数组         flow+=dfs(beg,INF,end);     }     return flow;//最大流量 } int main() {     int t;     scanf("%d",&t);     int k=1;     while(t--)     {         scanf("%d%d",&n,&m);         init();         getmap();         printf("Case %d: ",k++);         if(sum==Maxflow(0,sec))             printf("Yes\n\n");         else             printf("No\n\n");     }     return 0; }