POJ 1273 -Drainage Ditches

Time Limit:1000MS    Memory Limit:10000K

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

Source

USACO 93

题目大意：是说John有个池塘，每到下雨时节池塘里会被水草覆盖，附近有条河，于是他想挖沟渠放水，将水草清除掉。给你一组数据，数据包括沟渠的数目和每条沟渠的最大容量（及水的最大流量），还有沟渠的交汇点的数目。要你按照数据给的值求出从池塘流向河的最大流量。

很明显，这道题目就是求容量网络中的最大流...

代码：

 1 var
 2   c,e,n,m,i,s,t:longint;
 3   ans,inf:int64;
 4   h,d,f,g:array[0..5000]of longint;
 5   ot,cap,ne:array[0..100000]of longint;
 6
 7 procedure addedge(x,y,z:longint);
 8 begin
 9   ot[e]:=y; ne[e]:=g[x]; cap[e]:=z; g[x]:=e; inc(e);
10   ot[e]:=x; ne[e]:=g[y]; cap[e]:=0; g[y]:=e; inc(e);
11 end;
12
13 function min(a,b:int64):int64;
14 begin
15   if a<b then exit(a)
16          else exit(b);
17 end;
18
19 function bfs:boolean;
20 var
21   l,r,x,p:int64;
22 begin
23   for i:=1 to n do d[i]:=n+10;
24   l:=0; r:=1; h[1]:=s; d[s]:=0;
25   while l<r do
26     begin
27       inc(l);
28       p:=g[h[l]];
29       while p<>-1 do
30         begin
31           if (cap[p]<>0)and(d[ot[p]]>d[h[l]]+1) then
32             begin
33               inc(r);
34               h[r]:=ot[p];
35               d[ot[p]]:=d[h[l]]+1;
36             end;
37           p:=ne[p];
38         end;
39     end;
40   exit(d[t]<>n+10);
41 end;
42
43 function dfs(x,flow:int64):int64;
44 var
45   p,tmp:int64;
46 begin
47   if x=t then exit(flow);
48   p:=f[x]; dfs:=0;
49   while (p<>-1)and(dfs<flow) do
50     begin
51       if (cap[p]<>0)and(d[ot[p]]=d[x]+1) then
52         begin
53           tmp:=dfs(ot[p],min(flow-dfs,cap[p]));
54           dec(cap[p],tmp);
55           inc(cap[p xor 1],tmp);
56           inc(dfs,tmp);
57         end;
58       p:=ne[p];
59     end;
60   f[x]:=p;
61 end;
62
63 begin
64   inf:=high(int64);
65   while not eof do begin
66   readln(m,n);
67   e:=0;
68   fillchar(g,sizeof(g),255);
69   for i:=1 to m do
70     begin
71       readln(s,t,c);
72       addedge(s,t,c);
73     end;
74   s:=1; t:=n; ans:=0;
75   while bfs do
76     begin
77       for i:=1 to n do
78         f[i]:=g[i];
79       inc(ans,dfs(s,inf));
80     end;
81   writeln(ans); end;
82 end.