#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;
}
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。