Sap算法模板

最新推荐文章于 2020-02-07 05:33:37 发布

原创最新推荐文章于 2020-02-07 05:33:37 发布 · 424 阅读

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本文介绍了一种高效的求解最大流问题的SAP算法。通过详细解释算法流程，包括初始化网络流、读取图数据及执行增广路径寻找等步骤，为读者提供了深入理解该算法的基础。同时展示了关键的数据结构如边集、距离标号数组等。

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#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int MAXN = 210;
const int MAXM = 210*210;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int v, next;
	int f;
}edge[MAXM];

int cnt;
int n, m;

int first[MAXN];
int q[MAXN];

int gap[MAXN], h[MAXN], cur[MAXN], pre[MAXN];
//gap:优化统计高度数量数组，h: 距离标号数组；cur：当前弧数组；pre：前驱数组
 
void init()
{
	cnt = 0;
	memset(first, -1, sizeof(first));
	memset(gap, 0, sizeof(gap));
	memset(h, 0, sizeof(h));
	memset(pre, -1, sizeof(pre));
}

void init_flow()
{
	for(int i = 0; i < cnt; i += 2) //初始网络流 
	{
		edge[i].f += edge[i^1].f;
		edge[i^1].f = 0;
	}
}

void read_graph(int u, int v, int f)
{
	edge[cnt].v = v, edge[cnt].f = f;
	edge[cnt].next = first[u], first[u] = cnt++;
	edge[cnt].v = u, edge[cnt].f = 0; //反向弧流量为0 
	edge[cnt].next = first[v], first[v] = cnt++;
}

int Sap(int s, int t, int n)
{
	int cur_flow, flow_ans = 0;
	int neck, u = s;
	memcpy(cur, first, sizeof(first));
	gap[0] = n;
	while(h[s] < n)
	{
		if(u == t)
		{
			cur_flow = INF;
			for(int i = s; i != t; i = edge[cur[i]].v)
			{
				int f = edge[cur[i]].f;
				if(cur_flow > f)
				{
					neck = i;
					cur_flow = f;
				}
			}
			for(int i = s; i != t; i = edge[cur[i]].v)
			{
				int e = cur[i];
				edge[e].f -= cur_flow;
				edge[e^1].f += cur_flow;
			}
			flow_ans += cur_flow;
			u = neck;
		}
		int e;
		for(e = cur[u]; e != -1; e = edge[e].next)
		{
			int v = edge[e].v, f = edge[e].f;
			if(h[u] == h[v] + 1 && f) break;
		}
		if(e != -1)
		{
			int v = edge[e].v;
			cur[u] = e;
			pre[v] = u;
			u = v;
		}
		else
		{
			int temp = n;
			if(--gap[h[u]] == 0) break;
			cur[u] = first[u];
			for(int e = first[u]; e != -1; e = edge[e].next)
			{
				int v = edge[e].v, f = edge[e].f;
				if(f) temp = min(temp, h[v]);
			}
			h[u] = temp + 1;
			++gap[h[u]];
			if(u != s) u = pre[u];
		}
	}
	return flow_ans;
}