POJ 1637 Sightseeing tour （SAP | Dinic 混合欧拉图的判断）

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Sightseeing tour Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6448 Accepted: 2654 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 Northwestern Europe 2003

　　
　　给出一张混合图（有有向边，也有无向边），判断是否存在欧拉回路。
　　首先是对图中的无向边随意定一个方向，然后统计每个点的入度（indeg）和出度（outdeg），如果（indeg - outdeg）是奇数的话，一定不存在欧拉回路；
　　如果所有点的入度和出度之差都是偶数，那么就开始网络流构图：
　　1，对于有向边，舍弃；对于无向边，就按照最开始指定的方向建权值为 1 的边；
　　2，对于入度小于出度的点，从源点连一条到它的边，权值为（outdeg - indeg）/2；出度小于入度的点，连一条它到汇点的权值为（indeg - outdeg）/2 的边；
　　构图完成，如果满流（求出的最大流值 == 和汇点所有连边的权值之和），那么存在欧拉回路，否则不存在。
　　另附一个讲解欧拉图不错的博客：http://www.cnblogs.com/destinydesigner/archive/2009/09/28/1575674.html
　　SAP果然快，0ms：



#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int VM=210;
const int EM=1010;
const int INF=0x3f3f3f3f;
struct Edge{
int to,nxt;
int cap;
}edge[EM<<1];
int n,m,cnt,head[VM];
int src,des,tot,sum,indeg[VM],outdeg[VM];
int dep[VM],gap[VM],cur[VM],aug[VM],pre[VM];
void addedge(int cu,int cv,int cw){
edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
head[cu]=cnt++;
edge[cnt].to=cu;    edge[cnt].cap=0;   edge[cnt].nxt=head[cv];
head[cv]=cnt++;
}
int SAP(int n){
int max_flow=0,u=src,v;
int id,mindep;
aug[src]=INF;
pre[src]=-1;
memset(dep,0,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0]=n;
for(int i=0;i<=n;i++)
cur=head; // 初始化当前弧为第一条弧
while(dep[src]<n){
int flag=0;
if(u==des){
max_flow+=aug[des];
for(v=pre[des];v!=-1;v=pre[v]){     // 路径回溯更新残留网络
id=cur[v];
edge[id].cap-=aug[des];
edge[id^1].cap+=aug[des];
aug[v]-=aug[des];   // 修改可增广量，以后会用到
if(edge[id].cap==0) // 不回退到源点，仅回退到容量为0的弧的弧尾
u=v;
}
}
for(int i=cur;i!=-1;i=edge.nxt){
v=edge.to;    // 从当前弧开始查找允许弧
if(edge.cap>0 && dep==dep[v]+1){  // 找到允许弧
flag=1;
pre[v]=u;
cur=i;
aug[v]=min(aug,edge.cap);
u=v;
break;
}
}
if(!flag){
if(--gap[dep]==0)    /* gap优化，层次树出现断层则结束算法 */
break;
mindep=n;
cur=head;
for(int i=head;i!=-1;i=edge.nxt){
v=edge.to;
if(edge.cap>0 && dep[v]<mindep){
mindep=dep[v];
cur=i;   // 修改标号的同时修改当前弧
                }
}
dep=mindep+1;
gap[dep]++;
if(u!=src)  // 回溯继续寻找允许弧
u=pre;
}
}
return max_flow;
}
void Init(){
cnt=0;
memset(head,-1,sizeof(head));
memset(indeg,0,sizeof(indeg));
memset(outdeg,0,sizeof(outdeg));
}
int main(){
//freopen("input.txt","r",stdin);
int t;
scanf("%d",&t);
while(t--){
Init();
scanf("%d%d",&n,&m);
int u,v,c;
for(int i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&c);
indeg[v]++;
outdeg++;
if(c==0)
addedge(u,v,1);
}
int flag=1;
for(int i=1;i<=n;i++)
if((indeg-outdeg)%2==1){
flag=0;
break;
}
if(!flag)
puts("impossible");
else{
sum=0;
src=0,  des=n+1;
for(int i=1;i<=n;i++){  //无向边建图,有向边舍弃
if(indeg<outdeg)
addedge(src,i,(outdeg-indeg)/2);
else if(indeg>outdeg){
addedge(i,des,(indeg-outdeg)/2);
sum+=(indeg-outdeg)/2;
}
}
int ans=SAP(des+1);
if(sum==ans)
puts("possible");
else
puts("impossible");
}
}
return 0;
}
　　



#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int VM=210;
const int EM=1010;
const int INF=0x3f3f3f3f;
struct Edge{
int u,v,nxt;
int cap;
}edge[EM<<1];
int n,m,cnt,head[VM];
int src,des,tot,sum,dep[VM],indeg[VM],outdeg[VM];
void addedge(int cu,int cv,int cw){
edge[cnt].u=cu;     edge[cnt].v=cv;     edge[cnt].cap=cw;
edge[cnt].nxt=head[cu];     head[cu]=cnt++;
edge[cnt].u=cv;     edge[cnt].v=cu;     edge[cnt].cap=0;
edge[cnt].nxt=head[cv];     head[cv]=cnt++;
}
void Init(){
cnt=0;
memset(head,-1,sizeof(head));
memset(indeg,0,sizeof(indeg));
memset(outdeg,0,sizeof(outdeg));
}
int BFS(){
queue<int> q;
while(!q.empty())
q.pop();
memset(dep,-1,sizeof(dep));
dep[src]=0;
q.push(src);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=head;i!=-1;i=edge.nxt){
int v=edge.v;
if(edge.cap>0 && dep[v]==-1){    //没有标记,且可行流大于0
dep[v]=dep+1;
q.push(v);
}
}
}
return dep[des]!=-1;    //汇点是否成功标号,也就是说是否找到增广路
}
int DFS(int u,int minx){
if(u==des)
return minx;
int tmp;
for(int i=head;i!=-1;i=edge.nxt){
int v=edge.v;
if(edge.cap>0 && dep[v]==dep+1 && (tmp=DFS(v,min(minx,edge.cap)))){
edge.cap-=tmp;
edge[i^1].cap+=tmp;
return tmp;
}
}
dep=-1;
return 0;
}
int Dinic(){
int ans=0,tmp;
while(BFS()){
while(1){
tmp=DFS(src,INF);
if(tmp==0)
break;
ans+=tmp;
}
}
return ans;
}
int main(){
//freopen("input.txt","r",stdin);
int t;
scanf("%d",&t);
while(t--){
Init();
scanf("%d%d",&n,&m);
int u,v,c;
for(int i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&c);
indeg[v]++;
outdeg++;
if(c==0)
addedge(u,v,1);
}
int flag=1;
for(int i=1;i<=n;i++)
if((indeg-outdeg)%2==1){
flag=0;
break;
}
if(!flag)
puts("impossible");
else{
sum=0;
src=0,  des=n+1;
for(int i=1;i<=n;i++){  //无向边建图,有向边舍弃
if(indeg<outdeg)
addedge(src,i,(outdeg-indeg)/2);
else if(indeg>outdeg){
addedge(i,des,(indeg-outdeg)/2);
sum+=(indeg-outdeg)/2;
}
}
int ans=Dinic();
if(sum==ans)
puts("possible");
else
puts("impossible");
}
}
return 0;
}