HDU-4292 Food 简单网络流

最新推荐文章于 2019-10-10 16:56:13 发布

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最新推荐文章于 2019-10-10 16:56:13 发布

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本文通过一个具体的实例，详细解析了如何使用网络流算法解决特定问题。从建立模型到实现代码，逐步介绍了源点到各个节点间的关系及流量限制，并最终求解最大流。

依据题中所给定的关系构图，直接网络流。

源点到食物，食物到人，人拆成两个点，流量为1，再从人到饮料。

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define INF 0x3fffffff
#define F(x) (x)
#define N(x) (205+(x))
#define CPN(x) (410+(x))
#define D(x) (615+(x))
using namespace std;

int n, f, d, head[1005], dis[1005], idx;
int que[1000], front, tail;

const int SS = 0, TT = 1000;

struct Edge {
    int v, cap, next;
}e[200000];

void addedge(int x, int v, int cap) {
    ++idx;
    e[idx].v = v, e[idx].cap = cap;
    e[idx].next = head[x], head[x] = idx;
    ++idx;
    e[idx].v = x, e[idx].cap = 0;
    e[idx].next = head[v], head[v] = idx;    
}

bool bfs() {
    int pos;
    memset(dis, 0xff, sizeof (dis));
    dis[SS] = front = tail = 0;
    que[++tail] = SS;
    while (front != tail) {
        pos = que[++front];
        for (int i = head[pos]; i != -1; i = e[i].next) {
            if (e[i].cap > 0 && dis[e[i].v] == -1) {
                dis[e[i].v] = dis[pos] + 1;
                if (e[i].v == TT) return true;
                que[++tail] = e[i].v;
            }    
        }
    }
    return false;
}

int dfs(int u, int flow) {
    if (u == TT) {
        return flow;    
    }
    int f, tf = 0;
    for (int i = head[u]; i != -1; i = e[i].next) {
        if (dis[u]+1 == dis[e[i].v] && e[i].cap > 0 && (f = dfs(e[i].v, min(e[i].cap, flow - tf)))) {
            e[i].cap -= f;
            e[i^1].cap += f;
            tf += f;
            if (tf == flow) return flow;
        }
     }
     if (!tf) dis[u] = -1;
     return tf;
}

int dinic() {
    int ret = 0;
    while (bfs()) {
        ret += dfs(SS, INF);    
    }
    return ret;
}

int main() {
    int c;
    char like[205];
    while (scanf("%d %d %d", &n, &f, &d) == 3) {
        memset(head, 0xff, sizeof (head));
        idx = -1;
        for (int i = 1; i <= f; ++i) {
            scanf("%d", &c);    
            addedge(SS, F(i), c);
        }
        for (int i = 1; i <= d; ++i) {
            scanf("%d", &c);
            addedge(D(i), TT, c);
        }
        for (int i = 1; i <= n; ++i) {
            addedge(N(i), CPN(i), 1);
            scanf("%s", like+1);
            for (int j = 1; j <= f; ++j) {
                if (like[j] == 'Y') {
                    addedge(F(j), N(i), 1);    
                }
            }
        }
        for (int i = 1; i <= n; ++i) {
            scanf("%s", like+1);
            for (int j = 1; j <= d; ++j) {
                if (like[j] == 'Y') {
                    addedge(CPN(i), D(j), 1);    
                }    
            }    
        }
        printf("%d\n", dinic());
    }
    return 0;    
}