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HDU 3572 Task Schedule
Task Schedule
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 357264-bit integer IO format: %I64d Java class name: Main
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
24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2
Sample Output
Case 1: Yes Case 2: Yes
Source
2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC
解题：最大流。。建图确实有点。。。一下子想不到。。。还以为要拆边。。。
好吧 源点到每个任务建边容量为pi，每个任务到期限内的各天连容量为1的边，每天到汇点连容量为m的边。。。
现在很好理解了，如果满流即存在可行方案。。
每天最多m台机器同时工作。。。每个任务需要工作pi天。。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 20000;18 struct arc{19 int to,flow,next;20 arc(int x = 0,int y = 0,int z = -1){21 to = x;22 flow = y;23 next = z;24 }25 };26 arc e[maxn*10];27 int head[maxn],d[maxn],cur[maxn];28 int tot,S,T,n,m;29 void add(int u,int v,int flow){30 e[tot] = arc(v,flow,head[u]);31 head[u] = tot++;32 e[tot] = arc(u,0,head[v]);33 head[v] = tot++;34 }35 bool bfs(){36 memset(d,-1,sizeof(d));37 queue<int>q;38 d[T] = 1;39 q.push(T);40 while(!q.empty()){41 int u = q.front();42 q.pop();43 for(int i = head[u]; ~i; i = e[i].next){44 if(e[i^1].flow && d[e[i].to] == -1){45 d[e[i].to] = d[u] + 1;46 q.push(e[i].to);47 }48 }49 }50 51 return d[S] > -1;52 }53 int dfs(int u,int low){54 if(u == T) return low;55 int tmp = 0,a;56 for(int &i = cur[u]; ~i; i = e[i].next){57 if(e[i].flow && d[e[i].to] + 1 == d[u]&&(a=dfs(e[i].to,min(e[i].flow,low)))){58 e[i].flow -= a;59 e[i^1].flow += a;60 tmp += a;61 low -= a;62 if(!low) break;63 }64 }65 if(!tmp) d[u] = -1;66 return tmp;67 }68 int dinic(){69 int ans = 0;70 while(bfs()){71 memcpy(cur,head,sizeof(head));72 ans += dfs(S,INF);73 }74 return ans;75 }76 int main() {77 int cs,pi,si,ei,cao = 1;78 scanf("%d",&cs);79 while(cs--){80 scanf("%d %d",&n,&m);81 int mx = 0,sum = 0;82 memset(head,-1,sizeof(head));83 S = tot = 0;84 for(int i = 1; i <= n; ++i){85 scanf("%d %d %d",&pi,&si,&ei);86 sum += pi;87 for(int k = si; k <= ei; ++k) add(i,n+k,1);88 add(S,i,pi);89 mx = max(mx,ei);90 }91 T = n + mx + 1;92 for(int i = n+1; i < T; ++i) add(i,T,m);93 printf("Case %d: %s\n\n",cao++,dinic() == sum?"Yes":"No");94 }95 return 0;96 }
HDU 3572 Task Schedule
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