最大流——EK&&Dinic
学习地方：http://www.wutianqi.com/?p=3107
EK模板:O(V*E^2)
临街矩阵实现：
int bfs(int s,int e){
int i;
queue<int>q;
CL(vt,false); CL(pre,-1);//初始化
for (i = 0; i <= n + 1; ++i) f[i] = inf;//注意这里的范围
q.push(s);
vt[s] = true;
while (!q.empty()){
int u = q.front(); q.pop();
for (i = 0; i <= n + 1; ++i){
//如果回到父亲节点，或者该点已经被访问过，或者这条路不满足有流量
if (i == u || vt[i] || pre[i] != -1 || mp[u][i] <= 0) continue;
vt[i] = true;
f[i] = min(f[u],mp[u][i]);//选出该路径上的最小流量
pre[i] = u;//记录i的父亲为u
q.push(i);
if (i == e){
flag = true;
break;
}
}
if (flag) break;
}
if (f[e] == inf) return -1;
else return f[e];
}
int EK(int s,int e){
int i;
int ans = 0;
while (1){
flag = false;
int d = bfs(s,e);//寻找可行流
if (d == -1) break;
//更新流量
for (i = e; i != s; i = pre[i]){
int u = pre[i];
mp[u][i] -= d;
mp[i][u] += d;
}
ans += d;
}
return ans;
}
邻接表实现：
struct node{
int v,w;
int next;
}g[M*4 + 10];
int head[N],ct;
void init(){
ct = 0;
CL(head,-1);
}
void add(int u,int v,int w){
g[ct].v = v;
g[ct].w = w;
g[ct].next = head[u];
head[u] = ct++;
g[ct].v = u;
g[ct].w = 0;
g[ct].next = head[v];
head[v] = ct++;
}
int bfs(int s,int e){
int i;
for (i = 0; i <= n; ++i) f[i] = inf;
CL(vt,false); CL(pre,-1); CL(path,-1);
queue<int>q;
q.push(s); vt[s] = true;
while (!q.empty()){
int u = q.front(); q.pop();
for (i = head[u]; i != -1; i = g[i].next){
int v = g[i].v;
if (vt[v] || g[i].w <= 0 || pre[g[i].v] != -1) continue;
pre[v] = u; path[v] = i; vt[v] = true;//多了一个path记录可行流这条路上的路的编号
f[v] = min(f[u],g[i].w);
if (v == e) break;
q.push(v);
}
}
if (f[e] == inf) return -1;
else return f[e];
}
int EK(int s,int e){
int i,ans = 0;
while (1){
int d = bfs(s,e);
if (d == -1) break;
int tp;
for (i = e; i != s; i = pre[i]){
tp = path[i];//去路径编号
g[tp].w -= d;
g[tp^1].w += d;
}
ans += d;
}
return ans;
}
还有一种写法就是记录逆向边，这里不用记录们因为当前变与逆变abs（差值）= 1，取抑或即可。另一种写法http://blog.sina.com.cn/s/blog_6635898a0100ly53.html
Dinic模板：
邻接表实现：
struct node{
int v,w;
int next;
}g[M*4 + 10];
int head[N],ct;
void init(){
ct = 0;
CL(head,-1);
}
void add(int u,int v,int w){
g[ct].v = v;
g[ct].w = w;
g[ct].next = head[u];
head[u] = ct++;
g[ct].v = u;
g[ct].w = 0;
g[ct].next = head[v];
head[v] = ct++;
}
//分层
bool layer(int s,int e){
int i;
CL(level,-1);
level[s] = 0;
int l,r;
l = r = 0;
q[r] = s;
while (l <= r){
int u = q[l++];
for (i = head[u]; i != -1; i = g[i].next){
int v = g[i].v;
if (level[v] == -1 && g[i].w > 0){
level[v] = level[u] + 1;
q[++r] = v;
if (v == e) return true;
}
}
}
return false;
}
int find(int s,int e){
int i;
int ans = 0;
int top = 1;
while (top){
int u = (top == 1 ? s : g[q[top - 1]].v);//如果没有变肯定是起点，否则就是上一个边终点
if (u == e){
int MIN = inf,pos;
//找出最小流量
for (i = 1; i < top; ++i){
int tp = q[i];//注意这里取边的编号
if (g[tp].w < MIN){
MIN = g[tp].w;
pos = i;
}
}
//更新容量
for (i = 1; i < top; ++i){
int tp = q[i];//注意这里取边的编号
g[tp].w -= MIN;
g[tp^1].w += MIN;
}
ans += MIN;
top = pos;
}
else{//找可行流
for (i = head[u]; i != -1; i = g[i].next){
int v = g[i].v;
if (g[i].w > 0 && level[v] == level[u] + 1){
q[top++] = i;
break;
}
}
if (i == -1){//如果u没有可走的子节点
top--;
level[u] = -1;
}
}
}
return ans;
}
int Dinic(int s,int e){
int ans = 0;
while (layer(s,e)) ans += find(s,e);//分层成功找可行流
return ans;
}
上边的Dinic在写最大权闭合图时，超时，后来又搜了一个新模板：
bool layer(int s,int e)
{
int i;
CL(level,-1);
level[s] = 0;
int l = 0, r= 0;
q[r] = s;
while (l <= r){
int u = q[l++];
for (i = head[u]; i != -1; i = g[i].next)
{
int v = g[i].v;
if (level[v] == -1 && g[i].w > 0)
{
level[v] = level[u] + 1;
q[++r] = v;
if (v == e) return true;
}
}
}
return false;
}
ll dinic(int s,int e){
ll ans = 0;
while (layer(s,e))
{
int top = 0,u = s,i;
for (i = s; i <= e; ++i) out[i] = head[i];
while (out[s] != -1)
{
if (u == e)
{
ll MIN = inf;
for (i = top - 1; i >= 0; --i)
{
MIN = min(MIN,g[q[i]].w);
}
for (i = top - 1; i >= 0; --i)
{
g[q[i]].w -= MIN;
g[q[i]^1].w += MIN;
if (g[q[i]].w == 0) top = i;
}
ans += MIN;
u = g[q[top]].u;
}
else if (out[u] != -1 && g[out[u]].w > 0 && level[u] + 1 == level[g[out[u]].v])
{
q[top++] = out[u];
u = g[out[u]].v;
}
else
{
while (top > 0 && u != s && out[u] == -1)
u = g[q[--top]].u;
out[u] = g[out[u]].next;
}
}
}
return ans;
}
