POJ 1273 Drainage Ditches 网络流 ISAP

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 41 2 401 4 202 4 202 3 303 4 10

Sample Output

50

Source

传送门

题目大意是，，很多水管，然后巴拉巴拉。。。

总之求最大流裸题！（$

不得不说很不喜欢poj，hdu这种，，，虽然有些题目只能在它们上面做。

首先poj不能用bits/stdc++.h这个头文件。。。每题CE似乎必然。

然后poj的题目都是多组测试数据。。。靠，有什么意义吗？？

事实上的话，，我这题真的脑餐好多地方。。。

改了很久！后来发现，，反向边的容量应该是0，，

还有就是更新最大流的时候手贱了，+=写成==。。笔记本的问题。。

网络流上，ISAP的时间复杂度最多O(V^2*E)。。实际上远不到这个把。。

由于这题范围小，边不多，其实是可以用EK水过去的。。

对于没有练过ISAP的我来说，，还是写一发把。。

自己看的模板是这个blog：

http://blog.csdn.net/u013368721/article/details/30490255

哦提一句，学会了一个新姿势。。

！x，x=0返回1.

~x，x>=0返回1.

//#include<bits/stdc++.h>#include<cstdio>#include<cstring> using namespace std;int read(){    int x=0,f=1;char ch=getchar();    while (ch<'0' || ch>'9'){if (ch=='-') f=-1;ch=getchar();}    while (ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}const int N=205;int n,m,Ecnt;int pre[N],cur[N],num[N],Q[N],d[N];struct Edge{int next,to,C;}E[N<<1]; int head[N];void add(int x,int y,int z){E[Ecnt].next=head[x];E[Ecnt].to=y;E[Ecnt].C=z;head[x]=Ecnt++;}void BFS(int source,int sink){memset(num,0,sizeof(num));memset(d,255,sizeof(d));d[sink]=0;num[0]=1;int Head=0,tail=0;Q[tail++]=sink;while (Head!=tail){int u=Q[Head++];for (int i=head[u];~i;i=E[i].next){int j=E[i].to;if (~d[j]) continue;d[j]=d[u]+1;num[d[j]]++;Q[tail++]=j;}}}int Isap(int source,int sink){memcpy(cur,head,sizeof(cur));BFS(source,sink);int flow=0,u;u=pre[source]=source;while (d[sink]<n+1){if (u==sink){int f=999999,neck;for (int i=source;i!=sink;i=E[cur[i]].to)if (f>E[cur[i]].C){f=E[cur[i]].C;neck=i;}for (int i=source;i!=sink;i=E[cur[i]].to)E[cur[i]].C-=f,E[cur[i]^1].C+=f;flow+=f;u=neck; }int j;for (j=cur[u];~j;j=E[j].next)if (d[E[j].to]+1==d[u] && E[j].C)break;if (~j){cur[u]=j;pre[E[j].to]=u;u=E[j].to;}else{num[d[u]]--;if (!num[d[u]]) break;int mind=n+1;for (int i=head[u];~i;i=E[i].next)if (E[i].C && mind>d[E[i].to]){cur[u]=i;mind=d[E[i].to];}d[u]=mind+1;num[d[u]]++;u=pre[u];}}return flow;}void solve(){memset(head,255,sizeof(head));int x,y,z;Ecnt=0;for (int i=0;i<m;i++){x=read(),y=read(),z=read();add(x,y,z);add(y,x,0); //opposite side}printf("%d\n",Isap(1,n));}int main(){while (~scanf("%d%d",&m,&n))solve();return 0;}