链接:

https://vjudge.net/problem/HDU-4292

题意:

  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.

思路:

最大流,建图,模板题.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL; const int MAXN = 200+10;
const int INF = 1e9; struct Edge
{
int from, to, cap;
};
vector<int> G[MAXN*4];
vector<Edge> edges;
int Dis[MAXN*4];
int Fo, Dr;
int n, f, d, s, t; void Init()
{
for (int i = s;i <= t;i++)
G[i].clear();
edges.clear();
} void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge{from, to, cap});
edges.push_back(Edge{to, from, 0});
G[from].push_back(edges.size()-2);
G[to].push_back(edges.size()-1);
} bool Bfs()
{
memset(Dis, -1, sizeof(Dis));
queue<int> que;
que.push(s);
Dis[s] = 0;
while (!que.empty())
{
int u = que.front();
que.pop();
for (int i = 0;i < G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (e.cap > 0 && Dis[e.to] == -1)
{
Dis[e.to] = Dis[u]+1;
que.push(e.to);
}
}
}
return Dis[t] != -1;
} int Dfs(int u, int flow)
{
if (u == t)
return flow;
int res = 0;
for (int i = 0;i < G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
{
int tmp = Dfs(e.to, min(flow, e.cap));
e.cap -= tmp;
flow -= tmp;
res += tmp;
edges[G[u][i]^1].cap += tmp;
if (flow == 0)
break;
}
}
if (res == 0)
Dis[u] = -1;
return res;
} int MaxFlow()
{
int res = 0;
while (Bfs())
res += Dfs(s, INF);
return res;
} int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
while (cin >> n >> f >> d)
{
s = 0, t = n*2+f+d+1;
Init();
for (int i = 1;i <= f;i++)
{
cin >> Fo;
AddEdge(0, 2*n+i, Fo);
}
for (int i = 1;i <= d;i++)
{
cin >> Dr;
AddEdge(2*n+f+i, t, Dr);
}
for (int i = 1;i <= n;i++)
AddEdge(i*2-1, i*2, 1);
char ok;
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= f;j++)
{
cin >> ok;
if (ok == 'Y')
AddEdge(2*n+j, 2*i-1, 1);
}
}
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= d;j++)
{
cin >> ok;
if (ok == 'Y')
AddEdge(2*i, 2*n+f+j, 1);
}
}
int res = MaxFlow();
cout << res << endl;
} return 0;
}

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