HDU1529-Casher Emploryment(最最...最经典的差分约束差分约束-最长路+将环变线)

A supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to fit its need. The supermarket manager has hired you to help him, solve his problem. The problem is that the supermarket needs different number of cashiers at different times of each day (for example, a few cashiers after midnight, and many in the afternoon) to provide good service to its customers, and he wants to hire the least number of cashiers for this job.

The manager has provided you with the least number of cashiers needed for every one-hour slot of the day. This data is given as R(0), R(1), ..., R(23): R(0) represents the least number of cashiers needed from midnight to 1:00 A.M., R(1) shows this number for duration of 1:00 A.M. to 2:00 A.M., and so on. Note that these numbers are the same every day. There are N qualified applicants for this job. Each applicant i works non-stop once each 24 hours in a shift of exactly 8 hours starting from a specified hour, say ti (0 <= ti <= 23), exactly from the start of the hour mentioned. That is, if the ith applicant is hired, he/she will work starting from ti o'clock sharp for 8 hours. Cashiers do not replace one another and work exactly as scheduled, and there are enough cash registers and counters for those who are hired.

You are to write a program to read the R(i) 's for i=0...23 and ti 's for i=1...N that are all, non-negative integer numbers and compute the least number of cashiers needed to be employed to meet the mentioned constraints. Note that there can be more cashiers than the least number needed for a specific slot.

Input

The first line of input is the number of test cases for this problem (at most 20). Each test case starts with 24 integer numbers representing the R(0), R(1), ..., R(23) in one line (R(i) can be at most 1000). Then there is N, number of applicants in another line (0 <= N <= 1000), after which come N lines each containing one ti (0 <= ti <= 23). There are no blank lines between test cases.

Output

For each test case, the output should be written in one line, which is the least number of cashiers needed.

If there is no solution for the test case, you should write No Solution for that case.

Sample Input

1

1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

5

0

23

22

1

10

Sample Output

觉得好难分析...Orz 呜呜呜~ 还是自己太菜了

题解：这一题，我们可以用dis[i]表示从0到i总共雇佣的的Casher的数量，num[i]表示在i时正在工作的数量；sum[i]表示从0到i的需要工作时间的总时间；所以dis[23]即为所求，但是我们不能将23与0直接相连，会出现负环，故加一个汇点MAX=24指向0点，边权为 -dis[23];其实sum的最大值就是1000，也就1000中情况，故我们将sum从0~n意义枚举带入差分方程，如果有满足条件最长路则有解；

1.从8点到23点之间，满足dis[i]-dis[i-8]>=a[i];因为在i时，i-8的Casher已经停止工作了；

2.从0到22都满足 0=< sum[i+1]-sum[i] <=num[i+1];

3.在0到7点，由于可能含有昨天的剩余，则num[i]-num[i+16]>=a[i]-sum;

4.dis[i]-dis[MAX]>=0;

参考代码为：

#include<bits/stdc++.h>

using namespace std;

const int MAX=24;

const int INF=0x3f3f3f3f;

int T,n,x,a[24],temp[10],tot,tot2,sum;

int vis[25],dis[25],num[25],cnt[25],first[25];

struct Edge{

	int to,w,net;

} edge[3000];

void addedge(int u,int v,int w)

{

	edge[tot].to=v;

	edge[tot].w =w;

	edge[tot].net=first[u];

	first[u]=tot++;

}

queue<int> q;

void Init()

{

	memset(vis,0,sizeof vis);

	memset(dis,-INF,sizeof dis);

	memset(cnt,0,sizeof cnt);

	while(!q.empty()) q.pop();

	dis[MAX]=0;

}

void build()

{

	for(int i=8;i<24;i++) addedge(i-8,i,a[i]);

	for(int i=0;i<23;i++) addedge(i,i+1,0);

	for(int i=0;i<23;i++) addedge(i+1,i,-num[i+1]);

	for(int i=0;i<24;i++) addedge(MAX,i,0);

	tot2=tot-1;

	addedge(0,MAX,-num[0]);

	for(int i=0;i<8;i++)

	{

		temp[i]=tot;

		addedge(i+16,i,a[i]-0);

	}

}

void SPFA()

{

	bool flag=false;

	while(!flag&&sum<=n)

	{

		flag=true;

		Init();

		for(int i=0;i<8;i++) edge[temp[i]].w=a[i]-sum;

		edge[tot2].w=sum;

		q.push(MAX); vis[MAX]=1; cnt[MAX]++;

		while(!q.empty())

		{

			int u=q.front();q.pop();vis[u]=0;

			for(int i=first[u];~i;i=edge[i].net)

			{

				if(dis[edge[i].to]<dis[u]+edge[i].w)

				{

					dis[edge[i].to]=dis[u]+edge[i].w;

					if(!vis[edge[i].to])

					{

						vis[edge[i].to]=1;

						q.push(edge[i].to);

						if(++cnt[edge[i].to]>tot)

						{

							sum++;

							flag=false;

							break;

						}

					}

				}

			}

			if(!flag) break;

		}

	}

	if(flag) cout<<sum<<endl;

	else cout<<"No Solution"<<endl;

}

int main()

{

	ios::sync_with_stdio(false);

	cin.tie(0);

	cin>>T;

	while(T--)

	{

		sum=tot=0;

		memset(first,-1,sizeof first);

		memset(num,0,sizeof num);

		for(int i=0;i<24;i++) cin>>a[i];

		cin>>n;

		for(int i=1;i<=n;i++)

		{

			cin>>x;

			num[x]++;

		}

		build();

		SPFA();

	}

	return 0;

}

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HDU1529-Casher Emploryment(最最...最经典的差分约束差分约束-最长路+将环变线)

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HDU1529-Casher Emploryment(最最...最经典的差分约束 差分约束-最长路+将环变线)

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HDU1529-Casher Emploryment(最最...最经典的差分约束差分约束-最长路+将环变线)