Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11635 Accepted Submission(s): 3533
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
一开始读错题， 以为每个任务都要连续完成。 然后gg
网络流 判断满流 建图 ： 一个源点0，一个汇点t = 500 + n + 1; 每天建一个点 1 - 500 ，每个人物建一个点 500 + 1 - 500 + n;
从源点连边到每个task ，权值 为pi 。 从每个task 连边都每一个可以处理它的day的节点，权值为1，从每天连边到t，权值为m。从0-t 跑最大流，判断是否等于所有p之和，等于则满足。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2000 + 100;
const int maxm = 1000000 + 100;
typedef pair<int,int> P;
int n,m;
int l[maxn];//记录层数
int h[maxn];//链式前向星
int cur[maxn];
int tot = 0;
struct edge
{
int to;
int c;
int next;
edge(int x = 0, int y = 0, int z = 0) : to(x), c(y), next(z) {}
}es[maxm*2];//记录边 注意是2倍
void add_edge(int u, int v, int c)
{
es[tot] = edge(v,c,h[u]);
h[u] = tot++;
}
bool bfs(int s, int t)
{
memset(l,0,sizeof(l));
l[s] = 1;
queue <int> q;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
//cout << u << " " <<l[u] << endl;
if(u == t) return true;
for(int i = h[u]; i != -1; i = es[i].next)
{
int v = es[i].to;
if(!l[v] && es[i].c) {l[v] = l[u] + 1; q.push(v);}
}
}
return false;
}
int dfs(int x, int t, int mf)
{
if(x == t) return mf;
int ret = 0;
for(int &i = cur[x]; i != -1; i = es[i].next)
{
// cout << x << " " << l[es[i].to] << " "<< "mf = " << mf << endl;
if(es[i].c && l[x] == l[es[i].to] - 1)
{
// cout << es[i].c << " "<< l[es[i].to] << endl;
int f = dfs(es[i].to,t,min(es[i].c,mf - ret));
es[i].c -= f;
es[i^1].c += f;
ret += f;
if(ret == mf) return ret;
}
}
return ret;
}
int dinic(int s, int t)
{
int ans = 0;
while(bfs(s,t)) {
for(int i =
0; i <= t; i++) cur[i] = h[i];
int f = dfs(s,t,INF);
ans += f;
//if(f == 0) return ans;
//cout << f << endl;
}
return ans;
}
int main()
{
int T;
scanf("%d",&T);
int TT = T;
while(T--)
{
scanf("%d%d",&n,&m);
tot = 0;
memset(h,-1,sizeof(h));
int t = 500 + n + 1;
int p,s,e;
int res = 0;
for(int i = 1; i <= 500; i++) {add_edge(i,t,m);add_edge(t,i,0);}
for(int i = 1; i <= n; i++)
{
scanf("%d%d%d",&p,&s,&e);
add_edge(0,500+i,p);
add_edge(500+i,0,0);
res += p;
for(int j = s; j <= e; j++)
{
add_edge(i+500,j,1);
add_edge(j,i+500,0);//增加反向边
}
}
int ans = dinic(0,t);
printf("Case %d: ",TT-T);
if(ans >= res) printf("Yes\n");
else printf("No\n");
printf("\n");
}
return 0;
}