hdu 1532 Drainage Ditches

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8552    Accepted Submission(s): 3981

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. 

 

Input

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.

 

Output

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

   
   
   
   

    
    
    
    50
   
   
   
   

 

从which feed out of the pond and into each other  看出是无线图,最大网络流.

#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

#define ll int 

#define N 100040
#define M 205000
#define inf 107374182
#define inf64 1152921504606846976
struct Edge{  
	ll from, to, cap, nex;  
}edge[M*2];//注意这个一定要够大 不然会re 还有反向弧  

ll head[N], edgenum;  
void add(ll u, ll v, ll cap, ll rw = 0){ //如果是有向边则：add(u,v,cap); 如果是无向边则：add(u,v,cap,cap); 
	Edge E = { u, v, cap, head[u]};  
	edge[ edgenum ] = E;  
	head[u] = edgenum ++;  

	Edge E2= { v, u, rw,  head[v]};  
	edge[ edgenum ] = E2;  
	head[v] = edgenum ++;  
}  
ll sign[N];  
bool BFS(ll from, ll to){  
	memset(sign, -1, sizeof(sign));  
	sign[from] = 0;  

	queue<ll>q;  
	q.push(from);  
	while( !q.empty() ){  
		int u = q.front(); q.pop();  
		for(ll i = head[u]; i!=-1; i = edge[i].nex)  
		{  
			ll v = edge[i].to;  
			if(sign[v]==-1 && edge[i].cap)  
			{  
				sign[v] = sign[u] + 1, q.push(v);  
				if(sign[to] != -1)return true;  
			}  
		}  
	}  
	return false;  
}  
ll Stack[N], top, cur[N];  
ll Dinic(ll from, ll to){
	ll ans = 0;  
	while( BFS(from, to) )  
	{  
		memcpy(cur, head, sizeof(head));  
		ll u = from;      top = 0;  
		while(1)  
		{  
			if(u == to)  
			{  
				ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边  
				for(ll i = 0; i < top; i++)  
					if(flow > edge[ Stack[i] ].cap)  
					{  
						flow = edge[Stack[i]].cap;  
						loc = i;  
					}  

					for(ll i = 0; i < top; i++)  
					{  
						edge[ Stack[i] ].cap -= flow;  
						edge[Stack[i]^1].cap += flow;  
					}  
					ans += flow;  
					top = loc;  
					u = edge[Stack[top]].from;  
			}  
			for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标  
				if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;  
			if(cur[u] != -1)  
			{  
				Stack[top++] = cur[u];  
				u = edge[ cur[u] ].to;  
			}  
			else  
			{  
				if( top == 0 )break;  
				sign[u] = -1;  
				u = edge[ Stack[--top] ].from;  
			}  
		}  
	}  
	return ans;  
}
void init(){memset(head,-1,sizeof head);edgenum = 0;}

int main()
{
	int n,m,u,v,w;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		while(n--)
		{
			scanf("%d%d%d",&u,&v,&w);
		    add(u,v,w);
		}
		printf("%d\n",Dinic(1,m));
	}
	return 0;
}