Maximum flow non-recursive dinic algorithm for hdu3572 water problem

Task Schedule

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 4624 Accepted Submission (s): 1516

Problem Descriptionour Geometry Princess XMM has stoped she study in computational geometry to concentrate on her newly op Ened Factory. Her factory had introduced M new machines in order to process the coming N tasks. For the i-th task, the factory have to start processing it on or after day Si, process it for Pi days, and finish the task Before or at day Ei. A machine can have work on one task at a time, and each task can be processed by at a time. However, a task can is interrupted and processed on different machines on different days.
Now she wonders whether he had a feasible schedule to finish all the tasks in time. She turns to the help.

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

You are given-N (n<=500) and M (m<=200) on the first line of all test case. Then on each of next N lines is three integers Pi, si and Ei (1<=pi, Si, ei<=500), which has the meaning described In the description. It is guaranteed, this in a feasible schedule every task, can be finished would be do 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 Input24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

Sample outputcase 1:yes Case 2:yes

#include <iostream>#include<queue>#include<cstring>#include<cstdio>#include<climits>#defineMax (A, b) a>b?a:b#defineMin (A, b) a<b?a:busing namespacestd;structedge{intS,t,f,next;} edge[1100000];inthead[1010];intcur[1010];intpre[1010];intstack[1100000];intent;intn,m,times,s,t;voidAddintStartintLastintf) {EDGE[ENT].S=start;edge[ent].t=last;edge[ent].f=f;edge[ent].next=head[start];head[start]=ent++; Edge[ent].s=last;edge[ent].t=start;edge[ent].f=0; edge[ent].next=head[last];head[last]=ent++;}BOOLBFsintSintT) {memset (PRE,-1,sizeof(pre)); Pre[s]=0; Queue<int>Q;    Q.push (S);  while(!Q.empty ()) {        inttemp=Q.front ();        Q.pop ();  for(inti=head[temp];i!=-1; i=Edge[i].next) {            intTemp2=edge[i].t; if(pre[temp2]==-1&&edge[i].f) {PRE[TEMP2]=pre[temp]+1;            Q.push (TEMP2); }        }    }    returnpre[t]!=-1;}intDinic (intStartintLast ) {    intflow=0, now;  while(BFS (start,last)) {inttop=0; memcpy (Cur,head,sizeof(head)); intu=start;  while(1)        {            if(U==last)//If the end of the endpoint is found, the intermediate path is processed and the flow is calculated            {                intminn=Int_max;  for(intI=0; i<top;i++)                {                    if(minn>edge[stack[i]].f) {Minn=edge[stack[i]].f; now=i; }} Flow+=Minn;  for(intI=0; i<top;i++) {edge[stack[i]].f-=Minn; Edge[stack[i]^1].f+=Minn; } Top=Now ; U=Edge[stack[top]].s; }             for(inti=cur[u];i!=-1; cur[u]=i=edge[i].next)//find the side that you can go from the U point                if(edge[i].f&&pre[edge[i].t]==pre[u]+1)                     Break; if(cur[u]==-1)//If a viable edge is not found from this point, mark the point and backtrack            {                if(top==0) Break; Pre[u]=-1; U=edge[stack[--Top]].            S }            Else//If you find it, continue running{stack[top++]=Cur[u]; U=edge[cur[u]].t; }        }    }    returnflow;}intMain () {scanf ("%d",&Times );  for(intcas=1; cas<=times;cas++) {ent=0; memset (Head,-1,sizeof(head)); intst,ed,lt; intmaxn=0; intsum=0; S=0; t=1001; scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++) {scanf ("%d%d%d",&lt,&st,&ed); MAXN=Max (maxn,ed);  for(intj=st;j<=ed;j++) Add (j,i+ -,1); Add (i+ -, T,LT); Sum+=lt; }         for(intI=1; i<=maxn;i++) Add (s,i,m); if(Dinic (s,t) ==sum) printf ("Case %d:yes\n", CAs); Elseprintf"Case %d:no\n", CAs); printf ("\ n"); }    return 0;}

Maximum flow non-recursive dinic algorithm for hdu3572 water problem