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
给N个任务，M台机器。每个任务有最早才能开始做的时间S，deadline E，和持续工作的时间P。每个任务可以分段进行,但是在同一时刻,一台机器最多只能执行一个任务. 问存不存在可行的工作时间。
由于时间<=500且每个任务都能断断续续的执行,那么我们把每一天时间作为一个节点来用网络流解决该题.
建图: 源点s(编号0), 时间1-500天编号为1到500, N个任务编号为500+1 到500+N, 汇点t(编号501+N).
源点s到每个任务i有边(s, i, Pi)
每一天到汇点有边(j, t, M) (其实这里的每一天不一定真要从1到500,只需要取那些被每个任务覆盖的每一天即可)
如果任务i能在第j天进行,那么有边(i, j, 1) 注意由于一个任务在一天最多只有1台机器执行,所以该边容量为1,不能为INF或M哦.
最后看最大流是否 == 所有任务所需要的总天数.
1 #include <bits/stdc++.h>
2
3 using namespace std;
4 const int maxn = 2000;
5 const int inf = 0x3f3f3f3f;
6 struct Edge
7 {
8 int from,to,cap,flow;
9 Edge (int f,int t,int c,int fl)
10 {
11 from=f,to=t,cap=c,flow=fl;
12 }
13 };
14 struct Dinic
15 {
16 int n,m,s,t;
17 vector <Edge> edge;
18 vector <int> G[maxn];//存图
19 bool vis[maxn];//标记每点是否vis过
20 int cur[maxn];//当前弧优化
21 int dep[maxn];//标记深度
22 void init(int n,int s,int t)//初始化
23 {
24 this->n=n;this->s=s;this->t=t;
25 edge.clear();
26 for (int i=0;i<n;++i) G[i].clear();
27 }
28 void addedge (int from,int to,int cap)//加边,单向边
29 {
30 edge.push_back(Edge(from,to,cap,0));
31 edge.push_back(Edge(to,from,0,0));
32 m=edge.size();
33 G[from].push_back(m-2);
34 G[to].push_back(m-1);
35 }
36 bool bfs ()
37 {
38 queue<int> q;
39 while (!q.empty()) q.pop();
40 memset(vis,false,sizeof vis);
41 vis[s]=true;
42 dep[s]=0;
43 q.push(s);
44 while (!q.empty()){
45 int u=q.front();
46 //printf("%d\n",u);
47 q.pop();
48 for (int i=0;i<G[u].size();++i){
49 Edge e=edge[G[u][i]];
50 int v=e.to;
51 if (!vis[v]&&e.cap>e.flow){
52 vis[v]=true;
53 dep[v]=dep[u]+1;
54 q.push(v);
55 }
56 }
57 }
58 return vis[t];
59 }
60 int dfs (int x,int mi)
61 {
62 if (x==t||mi==0) return mi;
63 int flow=0,f;
64 for (int &i=cur[x];i<G[x].size();++i){
65 Edge &e=edge[G[x][i]];
66 int y=e.to;
67 if (dep[y]==dep[x]+1&&(f=dfs(y,min(mi,e.cap-e.flow)))>0){
68 e.flow+=f;
69 edge[G[x][i]^1].flow-=f;
70 flow+=f;
71 mi-=f;
72 if (mi==0) break;
73 }
74 }
75 return flow;
76 }
77 int max_flow ()
78 {
79 int ans = 0;
80 while (bfs()){
81 memset(cur,0,sizeof cur);
82 ans+=dfs(s,inf);
83 }
84 return ans;
85 }
86 }dinic;
87 int full_flow;
88 int main()
89 {
90 int casee = 0;
91 //freopen("de.txt","r",stdin);
92 int T;scanf("%d",&T);
93 while (T--){
94 int n,m;
95 full_flow = 0;
96 scanf("%d%d",&n,&m);
97 int src = 0,dst = 500+1+n;
98 dinic.init(500+2+n,src,dst);
99 bool vis[maxn];
100 memset(vis,false,sizeof vis);
101 for (int i=1;i<=n;++i){
102 int p,s,e;
103 scanf("%d%d%d",&p,&s,&e);
104 full_flow+=p;
105 dinic.addedge(src,i+500,p);
106 for (int j=s;j<=e;++j){
107 vis[j]=true;
108 dinic.addedge(i+500,j,1);
109 }
110 }
111 for (int i=0;i<maxn;++i){
112 if (vis[i])
113 dinic.addedge(i,dst,m);
114 }
115 printf("Case %d: ",++casee);
116 if (dinic.max_flow()==full_flow){//dinic.max_flow()只能跑一遍
117 printf("Yes\n\n");
118 }
119 else
120 printf("No\n\n");
121 }
122 return 0;
123 }