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最新推荐文章于 2024-09-20 15:39:23 发布

转载最新推荐文章于 2024-09-20 15:39:23 发布 · 104 阅读

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原文链接：https://my.oschina.net/dianpaopao/blog/165676

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#python

本文详细介绍了如何使用迪杰斯特拉算法解决网络流问题，并通过实例代码展示了算法的具体实现过程。重点突出了算法的时间复杂度和优化策略。

2019独角兽企业重金招聘Python工程师标准>>>

#include<stdio.h>
#include<string.h>
#define MAXD 3000
#define MAXM 26000
#define INF 100000000
int N, SUM;
int first[MAXD], next[MAXM], u[MAXM], v[MAXM], flow[MAXM], e;
int s[MAXD], d[MAXD], q[MAXD], work[MAXD];
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
void add(int a, int b, int w)
{
    u[e] = a;
    v[e] = b;
    flow[e] = w;
    next[e] = first[a];
    first[a] = e;
    e ++;
}
int init()
{
    int i, j, k, a;
    if(scanf("%d", &N) != 1)
        return 0;
    memset(first, -1, sizeof(first));
    e = SUM = 0;
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= N; j ++)
        {
            scanf("%d", &a);
            SUM += a;
            if((i + j) % 2 == 0)
            {
                add(0, i * N + j, a);
                add(i * N + j, 0, 0);
                for(k = 0; k < 4; k ++)
                {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(x >= 1 && x <= N && y >= 1 && y <= N)
                    {
                        add(i * N + j, x * N + y, INF);
                        add(x * N + y, i * N + j, 0);
                    }
                }
            }
            else
            {
                add(i * N + j, 1 , a);
                add(1, i * N + j, 0);
            }
        }
    return 1;
}
int bfs()
{
    int i, j, rear;
    memset(d, -1, sizeof(d));
    d[0] = 0;
    rear = 0;
    q[rear ++] = 0;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(d[v[j]] == -1 && flow[j])
            {
                d[v[j]] = d[q[i]] + 1;
                if(v[j] == 1)
                    return 1;
                q[rear ++] = v[j];
            }
    return 0;
}
int dfs(int cur, int a)
{
    if(cur == 1)
        return a;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i], a < flow[i] ? a : flow[i]))
            {
                flow[i] -= t;
                flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic()
{
    int res = 0, t;
    while(bfs())
    {
        memcpy(work, first, sizeof(first));
        while(t = dfs(0, INF))
            res += t;
    }
    return res;
}
int main()
{
    while(init())
    {
        int res = dinic();
        printf("%d\n", SUM - res);
    }
    return 0;
}

转载于:https://my.oschina.net/dianpaopao/blog/165676