补一下昨天的博客 J

题目链接:https://vjudge.net/contest/299467#problem/J

这个题目是一个裸的最小割问题,就不多赘述了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
struct edge
{
    int u, v, c, f;
    edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {}
};
vector<edge>e;
vector<int>G[maxn];
int level[maxn];//BFS分层,表示每个点的层数
int iter[maxn];//当前弧优化
int m;
void init(int n)
{
    for (int i = 0; i <= n; i++)G[i].clear();
    e.clear();
}
void add(int u, int v, int c)
{
    e.push_back(edge(u, v, c, 0));
    e.push_back(edge(v, u, c, 0));
    m = e.size();
    G[u].push_back(m - 2);
    G[v].push_back(m - 1);
}
void BFS(int s)//预处理出level数组
//直接BFS到每个点
{
    memset(level, -1, sizeof(level));
    queue<int>q;
    level[s] = 0;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int v = 0; v < G[u].size(); v++)
        {
            edge& now = e[G[u][v]];
            if (now.c > now.f && level[now.v] < 0)
            {
                level[now.v] = level[u] + 1;
                q.push(now.v);
            }
        }
    }
}
ll dfs(int u, int t, int f)//DFS寻找增广路
{
    if (u == t)return f;//已经到达源点,返回流量f
    for (int &v = iter[u]; v < G[u].size(); v++)
        //这里用iter数组表示每个点目前的弧,这是为了防止在一次寻找增广路的时候,对一些边多次遍历
        //在每次找增广路的时候,数组要清空
    {
        edge &now = e[G[u][v]];
        if (now.c - now.f > 0 && level[u] < level[now.v])
            //now.c - now.f > 0表示这条路还未满
            //level[u] < level[now.v]表示这条路是最短路,一定到达下一层,这就是Dinic算法的思想
        {
            ll d = dfs(now.v, t, min(f, now.c - now.f));
            if (d > 0)
            {
                now.f += d;//正向边流量加d
                e[G[u][v] ^ 1].f -= d;
                //反向边减d,此处在存储边的时候两条反向边可以通过^操作直接找到
                return d;
            }
        }
    }
    return 0;
}
ll Maxflow(int s, int t)
{
    ll flow = 0;
    for (;;)
    {
        BFS(s);
        if (level[t] < 0)return flow;//残余网络中到达不了t,增广路不存在
        memset(iter, 0, sizeof(iter));//清空当前弧数组
        int f;//记录增广路的可增加的流量
        while ((f = dfs(s, t, INF)) > 0)
        {
            flow += f;
        }
    }
    return flow;
}

vector<pair<int, int>>vec;

void solve(int u)
{
    printf("u=%d
", u);
    for(int i=0;i<G[u].size();i++)
    {
        edge now = e[G[u][i]];
        edge non = e[G[u][i] ^ 1];
        if(now.c==now.f)
        {
            printf("now.u=%d now.v=%d
", now.u, now.v);
            vec.push_back(make_pair(now.u, now.v));
            solve(now.v);
        }
    }
}
int cx[maxn], cy[maxn];
int main()
{
    int n, m;
    while (scanf("%d%d",&n,&m)!=EOF&&(m+n))
    {
        init(n);
        int s = 1, t = 2;
        for(int i=1;i<=m;i++)
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z);
            cx[i] = x;
            cy[i] = y;
        }
        ll ans = Maxflow(s, t);
        for(int i=1;i<=m;i++)
        {
            if((level[cx[i]]!=-1&&level[cy[i]]==-1)||(level[cx[i]]==-1&&level[cy[i]]!=-1))
            {
                printf("%d %d
", cx[i], cy[i]);
            }
        }
        printf("
");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/EchoZQN/p/10829984.html