bzoj1458

最小流

我很清楚知道自己已经忘记最小流怎么写了

是一个最小流,每行向每列连容量为1的边,源和汇分别向行和列连下界为限制,上界正无穷的边,然后就是跑最小流了。

我们先跑一个可行流,那么我们得先改造这个图,引入超级源汇,如果一个点流出去的比流进来的多,那么连向超级汇,否则超级源连过来,跑一遍dinic,这样得出了一个可行流,然后把汇向源连边,容量为inf,然后再跑一遍dinic,注意都是从超级源跑到超级汇,然后这条新建的边的反向边的流量就是最小流,原理不知道

#include<bits/stdc++.h>
using namespace std;
const int N = 310, inf = 1000000010;
int head[N], d[N], q[N], iter[N], Map[N][N];
struct edge {
    int nxt, to, f;
} e[N * N << 2];
int n, cnt = 1, source, sink, ans, m, k, sum1, sum2; 
int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
void link(int u, int v, int f)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].f = f;
}
void insert(int u, int v, int f)
{
    link(u, v, f);
    link(v, u, 0);
}
bool bfs()
{
    queue<int> q;
    q.push(source);
    memset(d, 0, sizeof(d));
    d[source] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i; i = e[i].nxt) if(e[i].f && !d[e[i].to])
        {
            d[e[i].to] = d[u] + 1;
            q.push(e[i].to);
            if(e[i].to == sink) return true;
        }
    }
    return false;
}
int dfs(int u, int delta)
{
    if(u == sink || delta == 0) return delta;
    int ret = 0;
    for(int &i = iter[u]; i; i = e[i].nxt) if(e[i].f && d[e[i].to] == d[u] + 1)
    {
        int x = dfs(e[i].to, min(e[i].f, delta));
        if(x == 0) d[e[i].to] = 0;
        e[i].f -= x;
        e[i ^ 1].f += x;
        delta -= x;
        ret += x;
        if(delta == 0) return ret;
    }
    return ret;
}
int dinic()
{
    int ret = 0;
    while(bfs())
    {
        for(int i = 0; i <= sink; ++i) iter[i] = head[i];
        ret += dfs(source, inf); 
    }
    return ret;
}
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    source = n + m + 2;
    sink = source + 1;
    for(int i = 1; i <= n; ++i)
    {
        int flow;
        scanf("%d", &flow);
        insert(0, i, inf);
        insert(source, i, flow);
        sum1 += flow;
    }
    insert(0, sink, sum1);
    for(int i = 1; i <= m; ++i)
    {
        int flow;
        scanf("%d", &flow);
        insert(i + n, n + m + 1, inf);
        insert(i + n, sink, flow);
        sum2 += flow; 
    }
    insert(source, n + m + 1, sum2);
    while(k--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        Map[x][y] = 1;
    }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) if(!Map[i][j]) insert(i, j + n, 1);
    dinic();
    insert(n + m + 1, 0, inf);
    dinic();
    ans = e[cnt].f;
    if(ans < max(sum1, sum2)) puts("JIONG!");
    printf("%d\n", ans); 
    return 0;
}
View Code

 

posted @ 2017-09-20 23:18  19992147  阅读(119)  评论(0编辑  收藏  举报