POJ 1815 Friendship ★(字典序最小点割集)

【题意】给出一个无向图，和图中的两个点s，t。求至少去掉几个点后才能使得s和t不连通，输出这样的点集并使其字典序最大。 不错的题，有助于更好的理解最小割和求解最小割的方法~ 【思路】 问题模型很简单，就是无向图的点连通度，也就是最小点割集。麻烦之处在于需要使得点割集方案的字典序最大。这样的话通常的dfs划分点集的方法就行不通了。我们采取贪心的策略：枚举1~N的点，在残留网络中dfs检查其代表的边的两端点是否连通，如果不连通则该点可以为点割，那么我们就让他是点割，加入到答案中，然后删掉这条边更新最大流。重复这个过程直到扫描完所有点。 PS:很多人判断点是否可以是点割时是先删掉边然后判断流是否减小，我觉得这样是不是麻烦了？因为如果不减小的话（不是点割）还得把他还原，相当于重新构了两次图。  

#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 = 50005;
const int oo = 0x3fffffff;

/* Dinic-2.0-2013.07.21: adds template.  double & int 转换方便多了，也不易出错 ~*/
template 
struct Dinic{
    struct node{
        int u, v;
        T flow;
        int opp;
        int next;
    }arc[2*MAXE];
    int vn, en, head[MAXV];
    int cur[MAXV];
    int q[MAXV];
    int path[2*MAXE], top;
    int dep[MAXV];
    void init(int n){
        vn = n;
        en = 0;
        mem(head, -1);
    }
    void insert_flow(int u, int v, T flow){
        arc[en].u = u;
        arc[en].v = v;
        arc[en].flow = flow;
        arc[en].next = head[u];
        head[u] = en ++;

        arc[en].u = v;
        arc[en].v = u;
        arc[en].flow = 0;
        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;
    }
    T solve(int s, int t){
        T 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;
                    T 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[path[k]^1].flow += minflow;
                    maxflow += minflow;
                    top = mink;
                    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  dinic;

bool map[MAXV][MAXV];
int n, s, t;
bool vis[MAXV];
bool dfs(int u, int t){
    vis[u] = 1;
    if (u == t)
        return true;
    for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){
        if (dinic.arc[i].flow <= 0) continue;
        int v = dinic.arc[i].v;
        if (!vis[v]){
            if (dfs(v, t)){
                return true;
            }
        }
    }
    return false;
}
bool del[MAXV];
void update_flow(){
    dinic.init(2*n);
    for (int i = 1; i <= n; i ++){
        if (del[i]) continue;
        if (i != s && i != t)
            dinic.insert_flow(i, n+i, 1);
        else{
            dinic.insert_flow(i, n+i, oo);
        }
        for (int j = 1; j <= n; j ++){
            if (del[j]) continue;
            if (i != j && map[i][j] == 1){
                dinic.insert_flow(n+i, j, oo);
            }
        }
    }
    dinic.solve(s, t);
}
int main(){
	//freopen("test.in", "r", stdin);
	//freopen("test.out", "w", stdout);
    while(scanf("%d %d %d", &n, &s, &t) != EOF){
        dinic.init(2*n);
        bool if_answer = 1;
        for (int i = 1; i <= n; i ++){
            if (i != s && i != t)
                dinic.insert_flow(i, n+i, 1);
            else{
                dinic.insert_flow(i, n+i, oo);
            }
            for (int j = 1; j <= n; j ++){
                scanf("%d", &map[i][j]);
                if (i != j && map[i][j] == 1){
                    dinic.insert_flow(n+i, j, oo);
                }
                if ((i == s && j == t && map[i][j] == 1))
                    if_answer = 0;
            }
        }
        if (!if_answer){
            puts("NO ANSWER!");
            continue;
        }
        int res = dinic.solve(s, t);
        printf("%d
", res);
        vector  mincut;
        mem(del, false);
        if (res){
            for (int i = 1; i <= n; i ++){
                mem(vis, 0);
                if (!dfs(i, n+i)){
                    mincut.push_back(i);
                    del[i] = true;
                    update_flow();
                }
            }
            for (int i = 0; i < (int)mincut.size(); i ++){
                if (i == 0)
                    printf("%d", mincut[i]);
                else
                    printf(" %d", mincut[i]);
            }
            puts("");
        }
    }
	return 0;
}