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.
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
Author
allenlowesy
Source
2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC
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\n\n",cnt++);
114 else printf("Case %d: No\n\n",cnt++);
115 }
116 }