| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 7647 | Accepted: 2210 |
Description
Ikki is the king of a small country – Phoenix, Phoenix is so small that there is only one city that is responsible for the production of daily goods, and uses the road network to transport the goods to the capital. Ikki finds that the biggest problem in the country is that transportation speed is too slow.
Since Ikki was an ACM/ICPC contestant before, he realized that this, indeed, is a maximum flow problem. He coded a maximum flow program and found the answer. Not satisfied with the current status of the transportation speed, he wants to increase the transportation ability of the nation. The method is relatively simple, Ikki will reconstruct some roads in this transportation network, to make those roads afford higher capacity in transportation. But unfortunately, the country of Phoenix is not so rich in GDP that there is only enough money to rebuild one road. Ikki wants to find such roads that if reconstructed, the total capacity of transportation will increase.
He thought this problem for a loooong time but cannot get it. So he gave this problem to frkstyc, who put it in this POJ Monthly contest for you to solve. Can you solve it for Ikki?
Input
The input contains exactly one test case.
The first line of the test case contains two integers N, M (N ≤ 500, M ≤ 5,000) which represents the number of cities and roads in the country, Phoenix, respectively.
M lines follow, each line contains three integers a, b, c, which means that there is a road from city a to city b with a transportation capacity of c (0 ≤ a, b < n, c ≤ 100). All the roads are directed.
Cities are numbered from 0 to n − 1, the city which can product goods is numbered 0, and the capital is numbered n − 1.
Output
Sample Input
2 1 0 1 1
Sample Output
1
Source
题意：
给出一张有向图，求最小割的必须边的个数
分析：
最近学数学学的有点伤心...(智商被完虐...)所以我们还是来写萌萌哒的网络流...
怎样求最小割的必须边？？？考虑构成最小割集的边一定是满流的边，并且把这些边去掉之后S无法到达T，所以我们可以枚举满流边，判断这条边的两个点是否是属于不同的集合...需要注意的是一个点可能被一群满流边包围所以导致无法访问到，所以我们把属于S集合的点标为1，T集合的点标为2...需要注意的是bfs的时候一定是残留网络中的所有边而不是只有原图中的边...
代码：
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 #include<queue> 6 //by NeighThorn 7 #define inf 0x3f3f3f3f 8 using namespace std; 9 //大鹏一日同风起，扶摇直上九万里 10 11 const int maxn=500+5,maxm=5000*2+5; 12 13 int n,m,ans,cnt,S,T,hd[maxn],to[maxm],fl[maxm],nxt[maxm],pos[maxn],vis[maxn]; 14 15 inline void add(int s,int x,int y){ 16 fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++; 17 fl[cnt]=0;to[cnt]=x;nxt[cnt]=hd[y];hd[y]=cnt++; 18 } 19 20 inline bool bfs(void){ 21 memset(pos,-1,sizeof(pos)); 22 int head=0,tail=0,q[maxn]; 23 q[0]=S;pos[S]=0; 24 while(head<=tail){ 25 int top=q[head++]; 26 for(int i=hd[top];i!=-1;i=nxt[i]) 27 if(pos[to[i]]==-1&&fl[i]) 28 pos[to[i]]=pos[top]+1,q[++tail]=to[i]; 29 } 30 return pos[T]!=-1; 31 } 32 33 inline int find(int v,int f){ 34 if(v==T) 35 return f; 36 int res=0,t; 37 for(int i=hd[v];i!=-1&&f>res;i=nxt[i]) 38 if(pos[to[i]]==pos[v]+1&&fl[i]) 39 t=find(to[i],min(fl[i],f-res)),fl[i]-=t,fl[i^1]+=t,res+=t; 40 if(!res) 41 pos[v]=-1; 42 return res; 43 } 44 45 inline void dinic(void){ 46 while(bfs()) 47 while(find(S,inf)); 48 } 49 50 inline void BFS(void){ 51 memset(vis,0,sizeof(vis)); 52 queue<int> q;q.push(S);vis[S]=1; 53 while(!q.empty()){ 54 int top=q.front();q.pop(); 55 for(int i=hd[top];i!=-1;i=nxt[i]) 56 if(fl[i]&&!vis[to[i]]) 57 vis[to[i]]=1,q.push(to[i]); 58 } 59 q.push(T),vis[T]=2; 60 while(!q.empty()){ 61 int top=q.front();q.pop(); 62 for(int i=hd[top];i!=-1;i=nxt[i]) 63 if(fl[i^1]&&!vis[to[i]]) 64 vis[to[i]]=2,q.push(to[i]); 65 } 66 } 67 68 signed main(void){ 69 memset(hd,-1,sizeof(hd)); 70 memset(pos,-1,sizeof(pos)); 71 scanf("%d%d",&n,&m);ans=cnt=0; 72 for(int i=1,s,x,y;i<=m;i++) 73 scanf("%d%d%d",&x,&y,&s),x++,y++,add(s,x,y); 74 S=1,T=n;dinic();BFS(); 75 for(int i=1;i<=n;i++) 76 for(int j=hd[i];j!=-1;j=nxt[j]) 77 if((j&1)==0&&!fl[j]) 78 ans+=((vis[i]==vis[S]&&vis[to[j]]==vis[T])?1:0); 79 printf("%d ",ans); 80 return 0; 81 }//Cap ou pas cap. Cap.
By NeighThorn