POJ 1637 Sightseeing tour
Sightseeing tour
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 9276 | Accepted: 3924 |
Description
The city executive board in Lund wants to construct a sightseeing tour by bus in Lund, so that tourists can see every corner of the beautiful city. They want to construct the tour so that every street in the city is visited exactly once. The bus should also start and end at the same junction. As in any city, the streets are either one-way or two-way, traffic rules that must be obeyed by the tour bus. Help the executive board and determine if it's possible to construct a sightseeing tour under these constraints.
Input
On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two positive integers m and s, 1 <= m <= 200,1 <= s <= 1000 being the number of junctions and streets, respectively. The following s lines contain the streets. Each street is described with three integers, xi, yi, and di, 1 <= xi,yi <= m, 0 <= di <= 1, where xi and yi are the junctions connected by a street. If di=1, then the street is a one-way street (going from xi to yi), otherwise it's a two-way street. You may assume that there exists a junction from where all other junctions can be reached.
Output
For each scenario, output one line containing the text "possible" or "impossible", whether or not it's possible to construct a sightseeing tour.
Sample Input
4 5 8 2 1 0 1 3 0 4 1 1 1 5 0 5 4 1 3 4 0 4 2 1 2 2 0 4 4 1 2 1 2 3 0 3 4 0 1 4 1 3 3 1 2 0 2 3 0 3 2 0 3 4 1 2 0 2 3 1 1 2 0 3 2 0
Sample Output
possible impossible impossible possible
Source
题意：
给出一张混合图，询问这张图是否存在欧拉回路(经过每条边一次且仅一次)...
分析：
如果一个有向图存在欧拉回路满足的条件是所有的点in[i]==out[i]，所以我们可以先给无向边定向，然后记录每个点的入度和出度，如果存在某一个点的入度和出度差值为奇数，那么这张图一定不存在欧拉回路，因为我们如果要找欧拉回路一定是通过反向无向边来使得每个点的入度等于出度，而每反向一条边，它所连接的两个点的入度和出度差值都改变了2...
那么对于一个入度不等于出度的点，我们需要把和它相邻的abs(in[i]-out[i])/2条边反向，所以对于一个in[i]>out[i]的点我们从i向T连一条容量为(in[i]-out[i])/2的边，对于一个out[i]>in[i]的点我们从S向i连一条容量为(out[i]-in[i])/2的边，然后我们把对于每个无向边，按照初始的定向连边，有向边删去(因为有向边是不能反向的)...如果可以满流就代表当前图满足要求...
代码：
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 //by NeighThorn 7 #define inf 0x3f3f3f3f 8 using namespace std; 9 10 const int maxn=200+5,maxm=4000+5; 11 12 int n,m,S,T,cas,cnt,sum,flag,hd[maxn],fl[maxm],to[maxm],in[maxn],out[maxn],nxt[maxm],pos[maxn]; 13 14 inline bool bfs(void){ 15 memset(pos,-1,sizeof(pos)); 16 int head=0,tail=0,q[maxn]; 17 q[0]=S,pos[S]=0; 18 while(head<=tail){ 19 int top=q[head++]; 20 for(int i=hd[top];i!=-1;i=nxt[i]) 21 if(pos[to[i]]==-1&&fl[i]) 22 pos[to[i]]=pos[top]+1,q[++tail]=to[i]; 23 } 24 return pos[T]!=-1; 25 } 26 27 inline int find(int v,int f){ 28 if(v==T) 29 return f; 30 int res=0,t; 31 for(int i=hd[v];i!=-1&&f>res;i=nxt[i]) 32 if(pos[to[i]]==pos[v]+1&&fl[i]) 33 t=find(to[i],min(f-res,fl[i])),res+=t,fl[i]-=t,fl[i^1]+=t; 34 if(!res) 35 pos[v]=-1; 36 return res; 37 } 38 39 inline int dinic(void){ 40 int res=0,t; 41 while(bfs()) 42 while(t=find(S,inf)) 43 res+=t; 44 return res; 45 } 46 47 inline void add(int s,int x,int y){ 48 fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++; 49 fl[cnt]=0;to[cnt]=x;nxt[cnt]=hd[y];hd[y]=cnt++; 50 } 51 52 signed main(void){ 53 // freopen("in.txt","r",stdin); 54 scanf("%d",&cas); 55 while(cas--){ 56 flag=cnt=sum=0; 57 scanf("%d%d",&n,&m); 58 memset(in,0,sizeof(in)); 59 memset(hd,-1,sizeof(hd)); 60 memset(out,0,sizeof(out)); 61 for(int i=1,s,x,y;i<=m;i++){ 62 scanf("%d%d%d",&x,&y,&s); 63 if(s==0) 64 add(1,x,y); 65 in[y]++,out[x]++; 66 }S=0,T=n+1; 67 for(int i=1;i<=n&&!flag;i++){ 68 if(abs(in[i]-out[i])&1) 69 flag=1; 70 else if(in[i]>out[i]) 71 add((in[i]-out[i])>>1,i,T); 72 else if(out[i]>in[i]) 73 add((out[i]-in[i])>>1,S,i),sum+=(out[i]-in[i])>>1; 74 } 75 if(flag){ 76 puts("impossible");continue; 77 } 78 if(dinic()==sum) 79 puts("possible"); 80 else 81 puts("impossible"); 82 } 83 return 0; 84 }
By NeighThorn
