题目链接：http://poj.org/problem?id=1273

题目大意：

有一个水塘，一个大海，要把水塘里的水尽可能多的排到大海里。当然，要通过一些水渠，这些水渠有一定的容量限制。

问最多能有多少水能流到大海里。

解题思路：

最大流的入门题。看了一上午就做了出来，很简单其实。。。用的BFS和Ford_Fulkerson算法。网上资料很多，大概看看应该就理解了。网络流开始还是很是不难的。建图是关键，要学会一定的建图方法。。。

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;

#define N 210
#define MAX 1 << 25

int network[N][N];
bool visit[N];
int que[N], head, tail; //广搜队列
int pre[N]; //祖先（用于更新残余网络）
int flow[N]; //可行流量
int n, m;

int BFS() //增广路
{
	head = tail = 0;
	for(int i = 1; i <= m; ++i)
		pre[i] = -1;
	for(int i = 1; i <= m; ++i)
		flow[i] = MAX;
	memset(visit, false, sizeof(visit));

	visit[1] = true;
	que[tail++] = 1;
	while(head < tail)
	{
		int cur = que[head++];
		if(cur == m)
			break;
		for(int i = 2; i <= m; ++i)
		{
			if(!visit[i] && network[cur][i]) //增广路
			{
				visit[i] = true;
				flow[i] = min(flow[cur], network[cur][i]);
				pre[i] = cur;
				que[tail++] = i;
			}
		}
	}
	if(!visit[m])
		return -1;
	else
		return flow[m];
}

int Ford_Fulkerson()
{
	int maxflow = 0, nowflow;
	while((nowflow = BFS()) != -1)
	{
		maxflow += nowflow;
		int temp = m, res;
		while(temp != 1) //更新残余网络
		{
			res = pre[temp];
			network[res][temp] -= nowflow;
			network[temp][res] += nowflow;
			temp = res;
		}
	}
	return maxflow;
}

int main()
{
	int start, end, water;
	while(~scanf("%d%d", &n, &m))
	{
		memset(network, 0, sizeof(network));
		for(int i = 0; i < n; ++i)
		{
			scanf("%d%d%d", &start, &end, &water); //重边叠加~。~
			network[start][end] += water;
		}
		printf("%d\n", Ford_Fulkerson());
	}
	return 0;
}

SAP实现代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

#define N 210
#define MAX 1<<28
#define CLR(arr, what) memset(arr, what, sizeof(arr))

int maze[N][N];
int pre[N], cur[N];
int gap[N];
int dis[N];
int source, sink, nodenum;

int SAP(int s, int t, int n)
{
	CLR(gap, 0); CLR(cur, 0); CLR(dis, 0);
	int u = pre[s] = s, maxflow = 0, aug = MAX, v;
	gap[0] = n;
	while(dis[s] < n)
	{
		bool flag = false;
		for(v = cur[u]; v <= n; ++v) //寻找允许弧
		{
			if(maze[u][v] > 0 && dis[u] == dis[v] + 1)
			{
				flag = true;
				break;
			}
		}
		if(flag) //找到允许弧
		{
			pre[v] = u;
			cur[u] = v;
			aug = min(aug, maze[u][v]);
			u = v;
			if(v == t) //找到完整增广路
			{
				maxflow += aug;
				for(v = t; v != s; v = pre[v]) //更新残留网络
				{
					maze[pre[v]][v] -= aug; //正向边
					maze[v][pre[v]] += aug; //反向边
				}
				aug = MAX, u = s; //重新从源点寻找
			}
		}
		else //找不到允许弧
		{
			int mindis = n;
			for(v = 1; v <= n; ++v) //重新标号
			{
				if(maze[u][v] && mindis > dis[v])
				{
					cur[u] = v;
					mindis = dis[v];
				}
			}
			if(--gap[dis[u]] == 0) //更新断层 + 判断是否断层（间隙优化）
				break;
			gap[dis[u] = mindis + 1]++; //更新断层
			u = pre[u]; //当前弧优化
		}
	}
	return maxflow;
}

int main()
{
	int n, m;
	int start, end, cost;
	int answer;
	while(~scanf("%d%d", &n, &m))
	{
		CLR(maze, 0);
		source = 1, sink = m, nodenum = m;
		for(int i = 0; i < n; ++i)
		{
			scanf("%d%d%d", &start, &end, &cost);
			maze[start][end] += cost;
		}
		answer = SAP(source, sink, nodenum);
		printf("%d\n", answer);
	}
	return 0;
}