解题报告 之 HDU4292 Food

解题报告 之 HDU4292 Food

Problem Description

　　You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
　　The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
　　You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
　　Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

 

Input

　　There are several test cases.
　　For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
　　The second line contains F integers, the ith number of which denotes amount of representative food.
　　The third line contains D integers, the ith number of which denotes amount of representative drink.
　　Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
　　Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, please print a single line with one integer, the maximum number of people to be satisfied.

 

Sample Input

4 3 3 1 1 1 1 1 1 YYN NYY YNY YNY YNY YYN YYN NNY

 

Sample Output

3

题目大意：n个人，f种食物，d种饮料，分别给出每个人对每种食物和饮料是否接受，Y/N。最后问最多能满足多少个人的需求。

分析：典型最大流题。属于多源点和多汇点的题目，还存在拆点，并将两个限制条件放在拆的点左右两边。感觉建图比较容易出错。首先多源点汇点的处理办法是建立一个超级源点和汇点，将多源点转化为单源点。那么超级源点对每个源点的边最大负载应该是多大呢？此题应该就是食物数那么大，因为从某食物源点出发的流不会超过这个值。然后将每个人拆点为两个，边负载为1，再按照Y/N将食物源点与人连起来，负载定为1或INF都行，因为其他地方会限制。重点来了，将饮料放在汇点之前，因为食物不好与饮料直接相连。处理办法同刚才对称。那么最后就求超级源点到超级汇点的最大流即可。

上代码：

#include
#include
#include
#include

using namespace std;

const int MAXN = 1510;
const int MAXM = 200010;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int to, next;
	int cap;
};

Edge edge[MAXM];

int N, F, D;
int cnt, src, des;
//cnt表示一共有几条边
int head[MAXN];
//head[i]表示节点i的第一条边的编号
int level[MAXN];

void addedge(int from, int to, int cap)
{
	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[from];
	head[from] = cnt++;

	edge[cnt].to = from; 
	edge[cnt].cap = 0;  
	edge[cnt].next = head[to];
	head[to] = cnt++;
}

int bfs()
{
	queue q;
	while (!q.empty())
		q.pop();
	memset(level, -1, sizeof(level));
	level[src] = 0; //源点的深度为0
	q.push(src);

	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (edge[i].cap > 0 && level[v] == -1) //该边未访问过且该边有流量
			{
				level[v] = level[u] + 1;
				q.push(v);
			}
		}
	}
	return level[des] != -1; //返回是否还存在连通路
}

int dfs(int u, int f)
{
	if (u == des)
		return f;
	int tem;
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if (edge[i].cap > 0 && level[v] == level[u] + 1)
		{
			tem = dfs(v, min(f, edge[i].cap));
			if (tem > 0)
			{
				edge[i].cap -= tem;
				edge[i ^ 1].cap += tem; //更新边与反向边的流量
				return tem;
			}
		}
	}
	level[u] = -1; //表示已经走过该点
	return 0;
}

int Dinic()
{
	int ans = 0, tem;
	while (bfs())//如果还存在连通路
	{
		while (tem = dfs(src, INF)) //一直试图寻找增广路
		{
			ans += tem;
		}
	}
	return ans;
}

int main()
{

	//freopen("input.txt","r",stdin);

	char str[220];
	while (~scanf("%d%d%d", &N, &F, &D))
	{
		cnt = 0;
		memset(head, -1, sizeof(head));
		int f, d;
		src = 0, des = F + 2 * N + D + 1;
		for (int i = 1; i <= F; i++)
		{
			scanf("%d", &f);
			addedge(src, i, f);
		}
		for (int i = F + 2 * N + 1; i <= F + 2 * N + D; i++)
		{
			scanf("%d", &d);
			addedge(i, des, d);
		}
		for (int i = 1; i <= N; i++)
		{
			scanf("%s", str);
			for (int j = 0; j < F; j++)
			if (str[j] == 'Y')
				addedge(j + 1, F + i, 1);
		}
		for (int i = 1; i <= N; i++)
		{
			scanf("%s", str);
			for (int j = 0; j < D; j++)
			if (str[j] == 'Y')
				addedge(F + N + i, F + 2 * N + j + 1, 1);
		}
		for (int i = F + 1; i <= F + N; i++)
			addedge(i, i + N, 1);
		printf("%d\n", Dinic());
	}
	return 0;
}

这里几道题都是用的Dinic，感觉KP算法复杂度肯定比不上Dinic，然后SAP大家可以去找一个代码作为模板备着，感觉不是很好理解。。过段时间再来看看吧。