算法1

有源汇上下界最小费用可行流

时间复杂度

C++ 代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 310, M = (5000 + N * 4) * 2, INF = 0x3f3f3f3f;
int n;
int h[N], e[M], ne[M], f[M], w[M], idx;
int din[N], dout[N];
int S, T;
int incf[N], d[N], q[N], pre[N], hh, tt;
bool st[N];

void add(int a, int b, int c, int d)
{
    e[idx] = b, ne[idx] = h[a], f[idx] = c, w[idx] = d, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], f[idx] = 0, w[idx] = -d, h[b] = idx ++; 
}

bool spfa()
{
    memset(d, 0x3f, sizeof d);
    memset(incf, 0, sizeof incf);
    incf[S] = INF, st[S] = true;
    q[0] = S, d[S] = 0, hh = 0, tt = 1;
    while(hh != tt)
    {
        int t = q[hh ++ ];
        st[t] = false;
        if(hh == N) hh = 0;
        for(int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if(f[i] && d[j] > d[t] + w[i])
            {
                d[j] = d[t] + w[i];
                incf[j] = min(incf[t], f[i]);
                pre[j] = i;
                if(!st[j])
                {
                    q[tt ++ ] = j;
                    if(tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return incf[T] > 0;
}

int EK()
{
    int flow = 0, cost = 0;
    while(spfa())
    {
        int t = incf[T];
        flow += t;
        cost += t * d[T];
        for(int i = T; i != S; i = e[pre[i] ^ 1])
        {
            f[pre[i]] -= t;
            f[pre[i] ^ 1] += t;
        }
    }
    return cost;
}

int main()
{
    scanf("%d", &n);
    memset(h, -1, sizeof h);
    S = 0, T = n + 2;
    int tot = 0;
    for(int i = 1; i <= n; ++ i)
    {
        int k;
        scanf("%d", &k);
        if(!k)
        {
            add(i, n + 1, INF, 0);
            continue;
        }
        for(int j = 0; j < k; ++ j)
        {
            int v, c;
            scanf("%d %d", &v, &c);
            add(i, v, INF, c);
            din[v] ++, dout[i] ++;
            tot += c;
        }
    }
    add(n + 1, 1, INF, 0);
    for(int i = 2; i <= n; ++ i) add(i, n + 1, INF, 0);
    for(int i = 1; i <= n + 1; ++ i)
    {
        if(din[i] > dout[i])
            add(S, i, din[i] - dout[i], 0);
        else if(din[i] < dout[i])
            add(i, T, dout[i] - din[i], 0);
    }
    printf("%d", EK() + tot);
    return 0;
}