hdu 3572 Task Schedule(網路流 dinic演算法)

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Task ScheduleTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3412    Accepted Submission(s): 1197


Problem DescriptionOur 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.
 
InputOn 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.
 
OutputFor 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

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

 
Sample Output

Case 1: Yes   Case 2: Yes

 
Authorallenlowesy 
思路：建一個超級源點0，然後假設工作區間長度為T ,再建立[1，T]個點，源點到每個點的流量為M（每天只有M台機器工作），接著，把相應的工作日向後平移T 天，每個工作日到相應的[1,T]的流量為1，到終點的流量也為1. 

最後求最大流是否大於等於總總工作量就是了。

#include"stdio.h"#include"string.h"#include"queue"using namespace std;#define N 1005#define max(a,b) (a>b?a:b)#define min(a,b) (a<b?a:b)const int inf=0x7ffffff;int cnt,n,m,t;int head[N],q[N],dis[N];struct node{    int u,v,w,next;}map[N*N];void add(int u,int v,int w){    map[cnt].u=u;    map[cnt].v=v;    map[cnt].w=w;    map[cnt].next=head[u];    head[u]=cnt++;    map[cnt].u=v;    map[cnt].v=u;    map[cnt].w=0;    map[cnt].next=head[v];    head[v]=cnt++;}int bfs(){    int i,u,v,t1,t2;    memset(dis,0,sizeof(dis));    u=t1=t2=0;    dis[u]=1;    q[t1++]=u;    while(t2<t1)    {        u=q[t2++];        for(i=head[u];i!=-1;i=map[i].next)        {            v=map[i].v;            if(map[i].w&&!dis[v])            {                dis[v]=dis[u]+1;                if(v==t)                    return 1;                q[t1++]=v;            }        }    }    return 0;}int dfs(int s,int lim){    int i,tmp,v,cost=0;    if(s==t)        return lim;    for(i=head[s];i!=-1;i=map[i].next)    {        v=map[i].v;        if(map[i].w&&dis[s]==dis[v]-1)        {            tmp=dfs(v,min(lim-cost,map[i].w));            if(tmp>0)            {                map[i].w-=tmp;                map[i^1].w+=tmp;                cost+=tmp;                if(cost==lim)                    break;            }            else                dis[v]=-1;        }    }    return cost;}int dinic(){    int ans=0,s=0;    while(bfs())        ans+=dfs(s,inf);    //printf("%d\n",ans);    return ans;}int main(){    int i,j,T,sum,t1,t2,cas=1;    int s[505],e[505],p[505];    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        t1=N;t2=0;        sum=0;        for(i=1;i<=n;i++)        {            scanf("%d%d%d",&p[i],&s[i],&e[i]);            t1=min(t1,s[i]);            t2=max(t2,e[i]);            sum+=p[i];        }        cnt=0;        memset(head,-1,sizeof(head));        for(i=t1;i<=t2;i++)      //超級源點到一般源點的流量        {            add(0,i,m);        }        for(i=1;i<=n;i++)        {            for(j=s[i];j<=e[i];j++)            {                add(j,j+t2,1);                add(j+t2,2*t2,1);            }        }        t=t2*2;        if(sum<=dinic())            printf("Case %d: Yes\n\n",cas++);        else            printf("Case %d: No\n\n",cas++);    }    return 0;}