圆桌问题【网络流24题】【最大流Dinic】

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题目链接


  为了让每个不同地方来的代表都分开来,所以我们要这么建边,从起点到每个代表都是建一条边权为人数的路径,然后再由代表出发,对每个餐桌建立边权为1的流,再由餐桌链接上汇点,此时的流是餐桌的容纳人数。


#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxE = 81555, maxN = 427;
const int S = 0;
int M, N, T, head[maxN], cur[maxN], cnt;
struct Eddge
{
    int nex, to, val;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), val(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int val)
{
    edge[cnt] = Eddge(head[u], v, val);
    head[u] = cnt++;
}
int deep[maxN];
queue<int> Q;
bool bfs()
{
    while(!Q.empty()) Q.pop();  memset(deep, 0, sizeof(deep));  deep[S] = 1;    Q.push(S);
    while(!Q.empty())
    {
        int u = Q.front();  Q.pop();
        for(int i=head[u], v, val; ~i; i=edge[i].nex)
        {
            v = edge[i].to; val = edge[i].val;
            if(!deep[v] && val)
            {
                deep[v] = deep[u] + 1;
                Q.push(v);
            }
        }
    }
    return deep[T];
}
int dfs(int u, int flow)
{
    if(u == T) return flow;
    for(int &i=cur[u], v, val; ~i; i=edge[i].nex)
    {
        v = edge[i].to; val = edge[i].val;
        if(deep[v] == deep[u] + 1 && val)
        {
            int di = dfs(v, min(val, flow));
            if(di)
            {
                edge[i].val -= di;
                edge[i^1].val += di;
                return di;
            }
        }
    }
    return 0;
}
int Dinic()
{
    int ans = 0, tmp = 0;
    while(bfs())
    {
        for(int i=S; i<=T; i++) cur[i] = head[i];
        while((tmp = dfs(S, INF))) ans += tmp;
    }
    return ans;
}
inline void init()
{
    cnt = 0;    T = N + M + 1;
    memset(head, -1, sizeof(head));
}
int main()
{
    scanf("%d%d", &M, &N);
    init();
    int all = 0;
    for(int i=1, person; i<=M; i++) //每个代表的人数
    {
        scanf("%d", &person);   all += person;
        addEddge(S, i, person);
        addEddge(i, S, 0);
        for(int j=1; j<=N; j++)
        {
            addEddge(i, j + M, 1);
            addEddge(j + M, i, 0);
        }
    }
    for(int i=1, V; i<=N; i++)
    {
        scanf("%d", &V);
        addEddge(i + M, T, V);
        addEddge(T, i + M, 0);
    }
    bool flag = Dinic() == all;
    printf("%d\n", flag);
    if(!flag) return 0;
    for(int u=1; u<=M; u++)
    {
        for(int i=head[u], v, val; ~i; i=edge[i].nex)
        {
            v = edge[i].to; val = edge[i^1].val;
            if(v > S && v < T && val) printf("%d ", v - M);
        }
        printf("\n");
    }
    return 0;
}

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转载自blog.csdn.net/qq_41730082/article/details/88697643
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