Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3036 Accepted Submission(s): 1086
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.
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.
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.
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
Author
allenlowesy
Source
Recommend
构图完成后就是模板题了。
构图：
先建立一个超级源点，然后与每个任务连接，容量为pi。然后每个任务与对应的日期连接，容量为1,。最后天数和一个超级汇点连接，容量为m。
当计算出来的最大流maxflow==sum(pi)时可完成任务，否则不能。
参考： http://blog.csdn.net/luyuncheng/article/details/7944417
代码：
1 //78MS 2200K 2777 B G++ 2 #include<stdio.h> 3 #include<string.h> 4 #define N 1005 5 #define inf 0x7ffffff 6 struct node{ 7 int v,c,next; 8 }edge[N*N]; 9 10 int head[N],eid; 11 int gap[N]; 12 int d[N]; 13 int pre[N]; 14 int curnode[N]; 15 int n,m,source,sink,nn; 16 17 void addedge(int u,int v,int c) 18 { 19 edge[eid].v=v; 20 edge[eid].c=c; 21 edge[eid].next=head[u]; 22 head[u]=eid++; 23 edge[eid].v=u; 24 edge[eid].c=0; 25 edge[eid].next=head[v]; 26 head[v]=eid++; 27 } 28 29 int ISAP() 30 { 31 int flow=0; 32 memset(d,0,sizeof(d)); 33 memset(gap,0,sizeof(gap)); 34 memset(pre,-1,sizeof(pre)); 35 for(int i=1;i<=nn;i++) curnode[i]=head[i]; 36 gap[nn]=nn; 37 int v,u=source,neck; 38 39 while(d[source]<nn){ 40 41 if(u==sink){ //找到增广路径 42 int tflow=inf; 43 for(int i=source;i!=sink;i=edge[curnode[i]].v){ 44 if(tflow>edge[curnode[i]].c){ 45 neck=i; 46 tflow=edge[curnode[i]].c; 47 } 48 } 49 for(int i=source;i!=sink;i=edge[curnode[i]].v){ 50 int temp=curnode[i]; 51 edge[temp].c-=tflow; 52 edge[temp^1].c+=tflow; 53 } 54 55 flow+=tflow; 56 u=neck; 57 } 58 59 for(v=curnode[u];v!=-1;v=edge[v].next) //查找允许弧 60 if(edge[v].c>0 && d[u]==d[edge[v].v]+1) 61 break; 62 63 if(v!=-1){ //找到允许弧 64 curnode[u]=v; 65 pre[edge[v].v]=u; 66 u=edge[v].v; 67 68 }else{ //不存在允许弧，回退一步 69 70 if(--gap[d[u]]==0) break; //断层结束算法 71 72 curnode[u]=head[u]; 73 int temp=nn; 74 for(int i=head[u];i!=-1;i=edge[i].next) 75 if(edge[i].c>0 && temp>d[edge[i].v]) 76 temp=d[edge[i].v]; 77 d[u]=temp+1; 78 ++gap[d[u]]; 79 if(u!=source) u=pre[u]; 80 } 81 } 82 83 return flow; 84 } 85 86 int main(void) 87 { 88 int t,p,s,e; 89 int k=1; 90 scanf("%d",&t); 91 while(t--) 92 { 93 source=0; 94 scanf("%d%d",&n,&m); 95 memset(head,-1,sizeof(head)); 96 eid=0; 97 int sump=0; 98 int maxn=0; 99 //构图 100 for(int i=1;i<=n;i++){ 101 scanf("%d%d%d",&p,&s,&e); 102 sump+=p; 103 addedge(source,i,p); 104 for(int j=s;j<=e;j++) 105 addedge(i,n+j,1); 106 maxn=maxn>e?maxn:e; 107 } 108 sink=n+maxn+1; 109 nn=sink+1; 110 for(int i=1;i<=maxn;i++) 111 addedge(i+n,sink,m); 112 113 printf("Case %d: ",k++); 114 if(ISAP()==sump) puts("Yes "); 115 else puts("No "); 116 } 117 return 0; 118 }