POJ 3469 最小割 最大流

题意就是有n个模块，每个模块可以运行在两个核心上，A核心和B核心，相应的有一个花费，有一些模块如果不在一个核心上运行就会产生额外的花费

现在要求最小的花费是的所有模块都运行

建图如下：

每个模块点，源点与其连边，容量为A花费，在用其与汇点连边，容量为相应B花费

然后如果有某对模块之间不运行在一个核心上会产生额外的花费，就对这两点建双向边，容量都为那个额外的花费

然后就是最小割模型

为啥这样建图就行呢， 可以观察， 如果一对点，假设为u,v之间不运行在一个模块会产生花费

首先，每个点不是与源点的边就是与汇点的边在割中，假如我们的割是有(s, u) (v, t) ，那么显然(v,u)这条边必须割掉，否则 s->v->u->t构成一条路径

如果割是(s,u) (s,v) 那么u,v之间的双向边一条也不需要割掉。也就满足了题意

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 22222
#define MAXM 1111111
#define INF 100000007
using namespace std;
struct node
{
    int v;    // vtex
    int c;    // cacity
    int f;   // current f in this arc
    int next, r;
}edge[MAXM];
int dist[MAXN], nm[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{
    edge[e].v = y;
    edge[e].c = c;
    edge[e].f = 0;
    edge[e].r = e + 1;
    edge[e].next = head[x];
    head[x] = e++;
    edge[e].v = x;
    edge[e].c = 0;
    edge[e].f = 0;
    edge[e].r = e - 1;
    edge[e].next = head[y];
    head[y] = e++;
}
void rev_BFS()
{
    int Q[MAXN], h = 0, t = 0;
    for(int i = 1; i <= n; ++i)
    {
        dist[i] = MAXN;
        nm[i] = 0;
    }
    Q[t++] = des;
    dist[des] = 0;
    nm[0] = 1;
    while(h != t)
    {
        int v = Q[h++];
        for(int i = head[v]; i != -1; i = edge[i].next)
        {
            if(edge[edge[i].r].c == 0 || dist[edge[i].v] < MAXN)continue;
            dist[edge[i].v] = dist[v] + 1;
            ++nm[dist[edge[i].v]];
            Q[t++] = edge[i].v;
        }
    }
}
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
int maxflow()
{
    rev_BFS();
    int u;
    int total = 0;
    int cur[MAXN], rpath[MAXN];
    for(int i = 1; i <= n; ++i)cur[i] = head[i];
    u = src;
    while(dist[src] < n)
    {
        if(u == des)     // find an augmenting path
        {
            int tf = INF;
            for(int i = src; i != des; i = edge[cur[i]].v)
                tf = min(tf, edge[cur[i]].c);
            for(int i = src; i != des; i = edge[cur[i]].v)
            {
                edge[cur[i]].c -= tf;
                edge[edge[cur[i]].r].c += tf;
                edge[cur[i]].f += tf;
                edge[edge[cur[i]].r].f -= tf;
            }
            total += tf;
            u = src;
        }
        int i;
        for(i = cur[u]; i != -1; i = edge[i].next)
            if(edge[i].c > 0 && dist[u] == dist[edge[i].v] + 1)break;
        if(i != -1)     // find an admissible arc, then Advance
        {
            cur[u] = i;
            rpath[edge[i].v] = edge[i].r;
            u = edge[i].v;
        }
        else        // no admissible arc, then relabel this vtex
        {
            if(0 == (--nm[dist[u]]))break;    // GAP cut, Important!
            cur[u] = head[u];
            int mindist = n;
            for(int j = head[u]; j != -1; j = edge[j].next)
                if(edge[j].c > 0)mindist = min(mindist, dist[edge[j].v]);
            dist[u] = mindist + 1;
            ++nm[dist[u]];
            if(u != src)
                u = edge[rpath[u]].v;    // Backtrack
        }
    }
    return total;
}
int nt, m;
int main()
{
    int u, v, w, A, B;
    scanf("%d%d", &nt, &m);
    src = nt + 1;
    des = nt + 2;
    n = des;
    init();
    for(int i = 1; i <= nt; i++)
    {
        scanf("%d%d", &A, &B);
        add(src, i, A);
        add(i, des, B);
    }
    for(int i = 1; i <= m; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }
    printf("%d\n", maxflow());
    return 0;
}