hdu3572--Task Schedule(最大流+两种优化方法，dinic)

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3651    Accepted Submission(s): 1271

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
   
   
   
   
   
   
   
   

    
    
    
    最大流问题，题意给出n种机器，m个任务，给出每一个任务完成需要的时间p，要在s,e时间段内完成，问可不可以完成（每个任务可以分部分做）
   
   
   
   
   
   
   
   

    
    
    
    一天数来建图，天数1到500，源点到每一个点的容量为n，对于每一个任务与在s到e内的点连线，容量是1，任务与汇点连线，容量是任务完成需要的时间，这样求出的最大流如果等于所有任务的时间和，那么就可以完成，都则完不成。
   
   
   
   
   
   
   
   

    
    
    
    使用了dinic算法，使用bfs建立层次图，再使用dfs更新增广路，第一个优化在dfs中如果找不到最终的汇点，那么将那个节点的的层次值改为-1，也就是在层次图中删掉该点，第二个优化，使用dfs回溯，通过一次dfs将所有可以更新的增广路全部更新。
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
#define maxn 1100
#define INF 0x3f3f3f3f
struct edge{
    int v , w ;
    int next ;
} p[1100000] ;
int head[maxn] , cnt , l[maxn] ;
queue <int> q ;
void add(int u,int v,int w)
{
    p[cnt].v = v ; p[cnt].w = w ;
    p[cnt].next = head[u] ; head[u] = cnt++ ;
    p[cnt].v = u ; p[cnt].w = 0 ;
    p[cnt].next = head[v] ; head[v] = cnt++ ;
}
int bfs(int s,int t)
{
    int u , v , i ;
    memset(l,-1,sizeof(l));
    l[s] = 0 ;
    while( !q.empty() )
        q.pop();
    q.push(s) ;
    while( !q.empty() )
    {
        u = q.front();
        q.pop();
        for(i = head[u] ; i != -1 ; i = p[i].next)
        {
            v = p[i].v ;
            if( l[v] == -1 && p[i].w )
            {
                l[v] = l[u] + 1 ;
                q.push(v) ;
            }
        }
    }
    if( l[t] > 0 )
        return 1 ;
    return 0 ;
}
int dfs(int s,int t,int min1)
{
    if( s == t )
        return min1 ;
    int i , v , a , ans = 0 ;
    for(i = head[s] ; i != -1 ; i = p[i].next)
    {
        v = p[i].v ;
        if( l[v] == l[s] + 1 && p[i].w && (a = dfs(v,t,min(min1,p[i].w) ) ) )
        {
            p[i].w -= a ;
            p[i^1].w += a ;
            ans += a ;
            min1 -= a ;
            if( !min1 )
                break;
        }
    }
    if( ans )
        return ans ;
    l[s] = -1 ;
    return 0;
}
int main()
{
    int t , tt , n , m , i , j , num , max_flow ;
    scanf("%d", &t);
    for(tt = 1 ; tt <= t ; tt++)
    {
        printf("Case %d: ", tt);
        cnt = 0 ; num = 0 ; max_flow = 0 ;
        memset(head,-1,sizeof(head));
        scanf("%d %d", &m, &n);
        int pp , s , e ;
        for(i = 1 ; i <= m ; i++)
        {
            scanf("%d %d %d", &pp, &s, &e);
            for(j = s ; j <= e ; j++)
            {

                add(j,500+i,1);
            }
            add(500+i,1001,pp);
            num += pp ;
        }
        for(i = 1 ; i <= 500 ; i++)
            add(0,i,n);
        while( bfs(0,1001) )
        {
            while( int k = dfs(0,1001,INF) )
                max_flow += k ;
        }
        if( num == max_flow )
            printf("Yes\n");
        else
            printf("No\n");
        printf("\n");
    }
    return 0;
}