HDU Problem - 4292 Food（最大流，建边）

题目链接

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. �阤 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. �阤 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

AC

建边
1. 源点到食物，权值为数量
2. 食物到人，权值为inf
3. 人到(拆点人)，权值为1（保证不会浪费，同一个人得到多个）
4. 拆点人到饮料，权值为inf
5. 饮料到汇点，权值为数量

#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define N 162000
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;
int head[810], curedge[810], level[810];
int inf = 0x3f3f3f3f;
int n, f, d, cnt;
struct ac{int v, c, pre;
}edge[N];
void init() {cnt = 0;memset(head, -1, sizeof(head));memset(edge, 0, sizeof(edge));
}
void addedge(int u, int v, int c) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v); edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].pre = head[u];head[u] = cnt++;
}bool bfs(int s, int e) {queue<int> que;memset(level, 0, sizeof(level));level[s] = 1;que.push(s);while (!que.empty()) {int u = que.front();que.pop();for (int i = head[u]; i != -1; i = edge[i].pre) {if (level[edge[i].v]  ||  edge[i].c <= 0)   continue;level[edge[i].v] = level[u] + 1;que.push(edge[i].v);}}return level[e] != 0;
}
int dfs(int s, int e, int flow) {if (s == e) return flow;for (int &i = curedge[s]; i != -1; i = edge[i].pre) {if (level[edge[i].v] == level[s] + 1 && edge[i].c > 0) {int d = dfs(edge[i].v, e, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i ^ 1].c += d;return d;}}}return 0;
}
int Dinic(int s, int e) {int sum = 0;while (bfs(s, e)) {for (int i = 0; i <= e; ++i) {curedge[i] = head[i];}int d;while (d = dfs(s, e, inf)) {sum += d;}}return sum;
}
int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifwhile (scanf("%d%d%d", &n, &f, &d) != EOF) {init();int start = 0, end =  f + n + n + d + 1;// 源点到食物 for (int i = 1; i <= f; ++i) {int t;scanf("%d", &t);addedge(start, i, t);}// 饮料到汇点 for (int j = 1; j <= d; ++j) {int t;scanf("%d", &t);addedge(f + n * 2 + j, end, t);}getchar();// 食物到人 for (int i = 1; i <= n; ++i) {char c;for (int j = 1; j <= f; ++j) {scanf("%c", &c);if (c == 'Y') {addedge(j, f + i, inf);}   }getchar();}// 人到饮料 for (int i = 1; i <= n; ++i) {char c;for (int j = 1; j <= d; ++j) {scanf("%c", &c);if (c == 'Y')   {addedge(f + n + i, f + n + n + j, inf);}}getchar();}// 拆点的同一个人之间建边 for (int i = 1; i <= n; ++i) {addedge(f + i, f + n + i, 1);}int ans = Dinic(start, end);printf("%d\n", ans);}return 0;
}