UVa10249 - The Grand Dinner(为什么超时)

最新推荐文章于 2014-08-10 11:34:17 发布

原创最新推荐文章于 2014-08-10 11:34:17 发布 · 748 阅读

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本文介绍了一个基于Edmonds-Karp算法的实现案例，该算法用于解决最大流问题。通过具体的代码示例展示了如何构建图、添加边以及进行最大流计算，并提供了完整的主函数流程。

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#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 130;
const int INF = 1000000;

struct Edge
{
    int from, to, cap, flow;
};

int m, n;
int s, t;
vector<Edge> vEdge;
vector<int> adjList[N];
int d[N], p[N];

void addEdge(int from, int to, int cap);
void init();
int Edmonds_Karp();
void print();

int main()
{
    int total, nMax;
    int cap;

#ifndef ONLINE_JUDGE
    freopen("d:\\OJ\\uva_in.txt", "r", stdin);
#endif

    while (scanf("%d%d", &m, &n) == 2 && (m || n)) {
        s = 0; t = m + n + 1;

        init();
        total = 0;
        nMax = 0;
        for (int i = 1; i <= m; i++) {
            scanf("%d", &cap);
            if (cap > nMax) nMax = cap;
            total += cap;

            addEdge(s, i, cap);
        }

        for (int i = m + 1; i <= m + n; i++) {
            scanf("%d", &cap);
            for (int j = 1; j <= m; j++) {
                addEdge(j, i, 1);
            }

            addEdge(i, t, cap);
        }

        if (m == 0) {
            printf("0\n");
            continue;
        }

        if (nMax > n || n == 0) {
            printf("0\n");
            continue;
        }

        int ans = Edmonds_Karp();
        if (ans == total) {
            printf("1\n");
            print();
        } else {
            printf("0\n");
        }
    }
    return 0;
}

void init()
{
    vEdge.clear();

    for (int i = 0; i <= t; i++) {
        adjList[i].clear();
    }
}

void addEdge(int from, int to, int cap)
{
    vEdge.push_back((Edge){from, to, cap, 0});
    vEdge.push_back((Edge){to, from, 0, 0});

    int num = vEdge.size();
    adjList[from].push_back(num - 2);
    adjList[to].push_back(num - 1);
}

int Edmonds_Karp()
{
    int f = 0;


    for (;;) {
        memset(d, 0x00, sizeof(d));
        d[s] = INF;

        queue<int> q;
        q.push(s);

        while (!q.empty()) {
            int cur = q.front(); q.pop();
            bool flag = false;
            for (int i = 0; i < adjList[cur].size(); i++) {
                int edgeNo = adjList[cur][i];
                Edge &e = vEdge[edgeNo];
                if (!d[e.to] && e.cap > e.flow) {
                    q.push(e.to);
                    p[e.to] = edgeNo;
                    d[e.to] = min(d[cur], e.cap - e.flow);

                    if (e.to == t) {
                        flag = true;
                        break;
                    }
                }
            }
            if (flag) break;
        }

        if (d[t] == 0) break;

        for (int u = t; u != s; u = vEdge[p[u]].from) {
            int num = p[u];
            vEdge[num].flow += d[t];
            vEdge[num ^ 1].flow -= d[t];
        }

        f += d[t];
    }

    return f;
}

void print()
{
    bool first;

    for (int i = 1; i <= m; i++) {
        first = true;
        for (int j = 0; j < adjList[i].size(); j++) {
            int no = adjList[i][j];
            Edge &e = vEdge[no];
            if (e.flow > 0) {
                if (first) first = false;
                else printf(" ");

                printf("%d", e.to - m);
            }
        }
        printf("\n");
    }
}