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 100000007using 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;}