题意 
现在有n个任务，两个机器A和B，每个任务要么在A上完成，要么在B上完成，而且知道每个任务在A和B机器上完成所需要的费用。然后再给m行，每行 a，b，w三个数字。表示如果a任务和b任务不在同一个机器上工作的话，需要额外花费w。现在要求出完成所有任务最小的花费是多少。
思路 
上次做的构图题是基于割截断s->t流与题目联系的，而这道题的构图则是基于割把流网络的点划分成了S、T点集，并且不同点集间的边都是割边。这是目前我所接触到的最小割的建模类型。
回到本题构图：源点向任务连一条Ai容量的边，任务向汇点连一条Bi容量的边，如果两任务在不同Core上需要代价则连一条(i,j,cost)的无向边。
代码   
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 505;
const int MAXE = 20005;
const int oo = 0x3fffffff;
struct node{
    int u, v, flow;
    int opp;
    int next;
};
struct Dinic{
    node arc[MAXE];
    int vn, en, head[MAXV];     //vn点个数(包括源点汇点),en边个数
    int cur[MAXV];              //当前弧
    int q[MAXV];                //bfs建层次图时的队列
    int path[MAXE], top;        //存dfs当前最短路径的栈
    int dep[MAXV];              //各节点层次
    void init(int n){
        vn = n;
        en = 0;
        mem(head, -1);
    }
    void insert_flow(int u, int v, int flow){
        arc[en].u = u;
        arc[en].v = v;
        arc[en].flow = flow;
        arc[en].opp = en + 1;
        arc[en].next = head[u];
        head[u] = en ++;

        arc[en].u = v;
        arc[en].v = u;
        arc[en].flow = 0;       //反向弧
        arc[en].opp = en - 1;
        arc[en].next = head[v];
        head[v] = en ++;
    }
    bool bfs(int s, int t){
        mem(dep, -1);
        int lq = 0, rq = 1;
        dep[s] = 0;
        q[lq] = s;
        while(lq < rq){
            int u = q[lq ++];
            if (u == t){
                return true;
            }
            for (int i = head[u]; i != -1; i = arc[i].next){
                int v = arc[i].v;
                if (dep[v] == -1 && arc[i].flow > 0){
                    dep[v] = dep[u] + 1;
                    q[rq ++] = v;
                }
            }
        }
        return false;
    }
    int solve(int s, int t){
        int maxflow = 0;
        while(bfs(s, t)){
            int i, j;
            for (i = 1; i <= vn; i ++)  cur[i] = head[i];
            for (i = s, top = 0;;){
                if (i == t){
                    int mink;
                    int minflow = 0x3fffffff;
                    for (int k = 0; k < top; k ++)
                        if (minflow > arc[path[k]].flow){
                            minflow = arc[path[k]].flow;
                            mink = k;
                        }
                    for (int k = 0; k < top; k ++)
                        arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
                    maxflow += minflow;
                    top = mink;		//arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
                    i = arc[path[top]].u;
                }
                for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
                    int v = arc[j].v;
                    if (arc[j].flow && dep[v] == dep[i] + 1)
                        break;
                }
                if (j != -1){
                    path[top ++] = j;
                    i = arc[j].v;
                }
                else{
                    if (top == 0)   break;
                    dep[i] = -1;
                    i = arc[path[-- top]].u;
                }
            }
        }
        return maxflow;
    }
}dinic;
bool vis[MAXV];
bool reach(int u, int p){
    vis[u] = 1;
    if (u == p)
        return true;
    for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){
        if (i % 2 == 1) continue;
        int v = dinic.arc[i].v;
        if (vis[v] || dinic.arc[i].flow <= 0) continue;
        if (reach(v, p))
            return true;
    }
    return false;
}
int work(int n){
    vector  seg;
    for (int i = 0; i < dinic.en; i += 2){
        if (dinic.arc[i].flow == 0){
            mem(vis, 0);
            int u = dinic.arc[i].u;
            int v = dinic.arc[i].v;
            if (reach(n+1, u) && reach(v, n+2)){
                seg.push_back(i/2+1);
            }
        }
    }
    for (int i = 0; i < (int)seg.size(); i ++){
        if (i == 0)
            printf("%d", seg[i]);
        else
            printf(" %d", seg[i]);
    }
    puts("");
}
int main(){
	//freopen("test.in", "r", stdin);
	//freopen("test.out", "w", stdout);
    int n, m, l;
    while(scanf("%d %d %d", &n, &m, &l), n){
        dinic.init(n+m+2);
        for (int i = 0; i < l; i ++){
            int u,v,w;
            scanf("%d %d %d", &u, &v, &w);
            dinic.insert_flow(u==0?n+m+2:u, v==0?n+m+2:v, w);
        }
        for (int i = 1; i <= n; i ++){
            dinic.insert_flow(n+m+1, i, oo);
        }
        dinic.solve(n+m+1, n+m+2);
        work(n+m);
    }
	return 0;
}
举杯独醉，饮罢飞雪，茫然又一年岁。 ------AbandonZHANG