Drainage Ditches(HDOJ 1532)

  1. 1:Problem Description
  2. 2:Input
  3. 3:Output
  4. 4:Sample Input
  5. 5:Sample Output
  6. 6:My Code

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. 


The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch. 


For each case, output a single integer, the maximum rate at which water may emptied from the pond. 

Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output


My Code


#include <iostream>
#include <cstring>
#include <queue>
#define maxn 205
#define inf 0x3f3f3f3f
using namespace std;
int n, m; //n是点数,m是边数
struct edge
	int v, w, next;
} e[maxn << 1]; //有反向边,要双倍
int tot;
int head[maxn];
int cur[maxn]; //当前弧优化,记录当前的弧
void add(int u, int v, int w)
	e[++tot].v = v;
	e[tot].w = w; //w是权重,即该边的容量
	e[tot].next = head[u];
	head[u] = tot;
int dep[maxn]; //存当前点的深度
void init()
	memset(e, -1, sizeof(e));
	memset(head, -1, sizeof(head));
	tot = -1;
bool bfs(int s, int t)
	memset(dep, 0, sizeof(dep));
	queue<int> q;
	dep[s] = 1;					 //源点
	for (int i = 1; i <= n; i++) //当前弧优化,拷贝一份head
		cur[i] = head[i];
	while (!q.empty())
		s = q.front();
		for (int i = head[s]; ~i; i = e[i].next)
			if (e[i].w > 0 && !dep[e[i].v]) //容量为0的边流不过去,已经流过的也不流了
				dep[e[i].v] = dep[s] + 1;
				if (e[i].v == t) //到终点就可以结束了,和终点同一层或更深层的我们是不会流到汇点的,所以不用管
					return true;
	return false; //汇点都流不到,肯定无法增广了
int dfs(int s, int t, int flow) //s为当前点,t为汇点,flow是流入当前点的剩余流量
	if (s == t || !flow) //流入汇点的流量就是该条路的流量,剩余流量没了,也结束
		return flow;
	int ans = 0;
	for (int &i = cur[s]; ~i; i = e[i].next) //当前弧优化,&是引用,i就相当于就是cur[s]的别名,改i和改cur[s]是一样的
		if (e[i].w && dep[e[i].v] == dep[s] + 1) //如果该点能流到儿子节点,且儿子节点是该点的下一层深度的
			int res = dfs(e[i].v, t, min(flow, e[i].w)); //那就继续往下流
			if (!res)
				dep[e[i].v] = 0; //如果这条路死了,那下次就不往这条路流了
			e[i].w -= res;		 //这条路的容量减小
			e[i ^ 1].w += res;	 //i^1,在这里等价于i+1,反悔边容量增加
			ans += res;			 //答案增加
			flow -= res;		 //剩余流量减小
			if (!flow)			 //流量用完了返回
	return ans;

int dinic(int s, int t)
	int ans = 0;
	while (bfs(s, t))		   //每次都先bfs建分层图
		ans += dfs(s, t, inf); //然后dfs找增广路
	return ans;
int main()
	while (cin >> m >> n)
		init(); //初始化
		for (int i = 0; i < m; i++)
			int u, v, w;
			cin >> u >> v >> w;
			add(u, v, w);
			add(v, u, 0); //建立反悔边
		cout << dinic(1, n) << endl;
	return 0;
Title - Artist