Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4323 Accepted Submission(s): 1447
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
思路:
1.主要是建图,0作为源点,1到n设置为任务,0到任务的边权为任务所需的天数,n到n+最大截止日期（top）设置为每一天,任务到相关每一天的边权设置为1,n+top+1设置为汇点,每一天到汇点的边权设置为机器数m;
2.直接把0作为源点，1-500作为任务，501-1000作为天数，1001作为汇点，建图会稍微简单一点；
3.判满流；
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstring> 5 #include<vector> 6 #include<queue> 7 #include<algorithm> 8 using namespace std; 9 const int inf=0x3fffffff; 10 const int maxn=1007; 11 struct Edge 12 { 13 int from,to,cap,flow; 14 Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}; 15 }; 16 17 struct Dinic{ 18 int n,m,s,t; 19 vector<Edge> edges; 20 vector<int> G[maxn]; 21 bool vis[maxn]; 22 int d[maxn]; 23 int cur[maxn]; 24 void init() 25 { 26 for(int i=0;i<maxn;i++) 27 { 28 G[i].clear(); 29 } 30 edges.clear(); 31 } 32 void AddEdge(int from,int to,int cap) 33 { 34 edges.push_back(Edge(from,to,cap,0)); 35 edges.push_back(Edge(to,from,0,0)); 36 int mm=edges.size(); 37 G[from].push_back(mm-2); 38 G[to].push_back(mm-1); 39 } 40 bool BFS(){ 41 memset(vis,0,sizeof(vis)); 42 queue<int>Q; 43 Q.push(s); 44 d[s]=0; 45 vis[s]=1; 46 while(!Q.empty()){ 47 int x=Q.front();Q.pop(); 48 for(int i=0;i<G[x].size();i++){ 49 Edge &e=edges[G[x][i]]; 50 if(!vis[e.to]&&e.cap>e.flow){ 51 vis[e.to]=1; 52 d[e.to]=d[x]+1; 53 Q.push(e.to); 54 } 55 } 56 } 57 return vis[t]; 58 } 59 int DFS(int x,int a){ 60 if(x==t||a==0) return a; 61 int flow=0,f; 62 for(int &i=cur[x];i<G[x].size();i++){ 63 Edge &e=edges[G[x][i]]; 64 if(d[e.to]==d[x]+1&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){ 65 e.flow+=f; 66 edges[G[x][i]^1].flow-=f; 67 flow+=f; 68 a-=f; 69 if(a==0) break; 70 } 71 } 72 return flow; 73 } 74 int Maxflow(int s,int t){ 75 this->s=s;this->t=t; 76 int flow=0; 77 while(BFS()){ 78 memset(cur,0,sizeof(cur)); 79 flow+=DFS(s,inf); 80 } 81 return flow; 82 } 83 }; 84 Dinic dic; 85 int main() 86 { 87 // freopen("in.txt","r",stdin); 88 int tt,cnt=1;int n,s,a,b,c,t,m; 89 scanf("%d",&tt); 90 int top=-1; 91 int sum; 92 while(tt--){ 93 dic.init(); 94 sum=0; 95 scanf("%d%d",&n,&m); 96 for(int i=1;i<=n;i++) 97 { 98 scanf("%d%d%d",&a,&b,&c); 99 sum+=a; 100 if(c>top) top=c; 101 dic.AddEdge(0,i,a); 102 for(int j=b;j<=c;j++) 103 { 104 dic.AddEdge(i,n+j,1); 105 } 106 } 107 for(int i=n+1;i<=n+top;i++) 108 { 109 dic.AddEdge(i,n+top+1,m); 110 } 111 s=0,t=n+top+1; 112 int ans=dic.Maxflow(s,t); 113 if(ans==sum) printf("Case %d: Yes ",cnt++); 114 else printf("Case %d: No ",cnt++); 115 } 116 }