hdu 3572 Task Schedule ( 最大流 + 满流）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3572

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7440    Accepted Submission(s): 2307

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

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

 

Sample Output

Case 1: Yes   Case 2: Yes

 

Author

allenlowesy

 

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC

 

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题目大意：n个任务，m个机器，输入pi, si, ei 分别表示 第i个任务需要p天，从s到e的任意p天，求是否能完成

解析 ：       最大流 + 满流问题，先学会EK或者Dinic求最大流，该题关键是如何建图，以0为源点，1~n为任务，边权为流量（这里是天数p），然后再将任务与天数对应起来，其中流量为1，再将天数与汇点对应起来，流量为机器数m（一天有m个机器在工作），最后判断是否满流，最后注意用引用减少时间，不然LTE

                    借鉴大牛博客：http://blog.csdn.net/discreeter/article/details/49004837

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<vector>#include<set>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N 1009using namespace std;const int inf = 0x3f3f3f3f;const int mod = 1<<30;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long LL;typedef struct{    int to, c, rec;}Q;vector<Q> mp[N];int pre[N], tt;int vis[N];void add(int u, int v, int w){    Q e;    e.to = v; e.c = w; e.rec = mp[v].size();    mp[u].push_back(e);    e.to = u; e.c = 0; e.rec = mp[u].size() - 1;    mp[v].push_back(e);}int bfs(){    queue<int> q;    memset(pre, 0, sizeof(pre));    int i;    q.push(0);    pre[0] = 1;    while(!q.empty())    {        int u = q.front(); q.pop();        int k = mp[u].size();        for(i = 0; i < k; i++)        {            Q &e = mp[u][i];            if(!pre[e.to] && e.c > 0)            {                pre[e.to] = pre[u] + 1;                q.push(e.to);            }        }    }    return pre[tt] > 0;}int dfs(int u, int cur_flow){    if(u == tt) return cur_flow;    int k = mp[u].size();    for(int &i = vis[u]; i < k; i++)    {        Q &e = mp[u][i];        if(pre[u] + 1 == pre[e.to] && e.c > 0)        {            int f = dfs(e.to, min(cur_flow, e.c));            if(f > 0)            {                e.c -= f;                mp[e.to][e.rec].c += f;                return f;            }        }    }    return 0;}int Dinic(){    int ans = 0, f;     while(bfs())     {         memset(vis, 0, sizeof(vis)); //很重要，减少大量时间         while((f = dfs(0, inf)))            ans += f;     }     return ans;}int main(){    int i, j, s, e, p, m ,n;    int t, cnt = 0;    cin >> t;    while(t--)    {        scanf("%d%d", &n, &m);        int sum = 0; tt = 0;        for(i = 1; i <= n; i++)        {            scanf("%d%d%d", &p, &s, &e);            add(0, i, p);            sum += p;            tt = max(tt, e);            for(j = s; j <= e; j++) add(i, j + n, 1);        }        tt = tt + n + 1;        for(i = n + 1; i < tt; i++) add(i, tt, m);        int ans = Dinic();        if(sum == ans) printf("Case %d: Yes\n\n", ++cnt);        else printf("Case %d: No\n\n", ++cnt);        for(i = 0; i < N; i++) mp[i].clear();    }    return 0;}