Island Transport(网络流之最大流 dicnic优化算法)

dicnic 算法介绍：SYCstudio

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int cas = 1;
int dist[maxn];//分层图中标记深度
int cur[maxn];//cur就是记录当前点u循环到了哪一条边
int start, tend;//源点 汇点 

struct node
{
	int to, cap, next;
	//node(){}
	//node(int a,int b,int c):to(a),cap(b),rev(c){}
}edge[maxn << 1];
//vectormp[maxn];
int head[maxn];
bool vis[maxn];
int cnt;//边的数量，从开始编号 
void addedge(int start, int to, int cap)
{
	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[start];
	head[start] = cnt++;
}
bool BFS()
{
	//memset(vis,false,sizeof(vis));
	memset(dist, -1, sizeof(dist));
	int Q[maxn * 2];
	int Thead, Ttail;
	Thead = Ttail = 0;
	Q[Ttail++] = start;;
	dist[start] = 0;
	//vis[start]=1;
	while (Thead0)//没有标记,且可行流大于0
			{
				//vis[temp.to]=true;
				dist[temp] = dist[x] + 1;
				Q[Ttail++] = temp;
			}
		}
		Thead++;
	}
	return false;//汇点是否成功标号,也就是说是否找到增广路
}
int DFS(int x, int cap)
{
	if (x == tend)return cap;
	//vis[start]=true;
	int flow = 0, f;
	for (int &i = cur[x]/*head[x]*/; i != -1; i = edge[i].next)
	{
		int temp = edge[i].to;
		//if(temp.cap<=0||vis[temp.to])continue;
		if (dist[temp] == dist[x] + 1 && edge[i].cap)
		{
			f = DFS(temp, min(cap - flow, edge[i].cap));
			edge[i].cap -= f;
			edge[i ^ 1].cap += f;
			flow += f;
			if (flow == cap)break;
		}
	}
	if (!flow)dist[x] = -2;//防止重搜
	return flow;
}
int Dicnic(int n)
{
	int flow = 0, f;
	while (BFS())
	{
		for(int i=1;i<=n;i++)
		cur[i]=head[i];
		//memset(vis,false,sizeof(vis));
		while ((f = DFS(start, INF))>0)
			flow += f;
	}
	return flow;
}
int main()
{	
	int t;
	scanf("%d",&t);

	int n, m;

	while (	t--)
	{	
		scanf("%d%d", &n, &m);
		
		cnt = 0;
		memset(head, -1, sizeof(head));
		int x, y, u, v, w;
		//for(int i=0;i=tmax){
				tmax=x;
				tend=i;
			}
			if(x<=tmin){
				tmin=x;
				start=i;
			}
		}
		while (m--)
		{
			scanf("%d%d%d", &u, &v, &w);
			addedge(u, v, w);
			addedge(v,u,w);//双向边，单向时，w=0 
		}
		int ans = Dicnic(n);//n为点的个数 
		printf("%d\n", ans);
	}
}