Sightseeing tour
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 9100 | Accepted: 3830 |
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
【分析】
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（不一定是1，应该是这条边出现的次数，因为可能重边，我就是在这个地方WA了）；
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。
#include <iostream> #include <cstring> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #define inf 0x3f3f3f3f #define mod 10000 typedef long long ll; using namespace std; const int N=405; const int M=300005; int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;} struct man { int c,f; }w[N][N]; int dis[N],n,m; int t,cnt,maxn,ans; int in[N],out[N]; bool flag; bool bfs() { queue<int>q; memset(dis,0,sizeof(dis)); q.push(0); dis[0]=1; while(!q.empty() && !dis[t]){ int v=q.front();q.pop(); for(int i=1;i<=t;i++){ //if(i==t)printf("w[i][t].c=%d ",w[i][t].c); if(!dis[i]&&w[v][i].c>w[v][i].f){ q.push(i); dis[i]=dis[v]+1; } } } return dis[t]!=0; } int dfs(int cur,int cp) { if(cur==t||cp==0)return cp; int tmp=cp,tt; for(int i=1;i<=t;i++){ if(dis[i]==dis[cur]+1 &&w[cur][i].c>w[cur][i].f){ tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp)); w[cur][i].f+=tt; w[i][cur].f-=tt; tmp-=tt; } } return cp-tmp; } void dinic() { ans=0; while(bfs())ans+=dfs(0,inf); if(ans==maxn)puts("possible"); else puts("impossible"); } void init() { int a,b,d; scanf("%d%d",&n,&m);t=n+1; for(int i=1;i<=n;i++)in[i]=out[i]=0; while(m--){ scanf("%d%d%d",&a,&b,&d); if(d==0)w[a][b].c++;//有重边，若把它赋值为1，WA out[a]++;in[b]++; } } void solve() { flag=true; for(int i=1;i<=n;i++){ if((in[i]-out[i])&1){ puts("impossible");flag=false;return; } if(in[i]<out[i])w[0][i].c=(out[i]-in[i])/2; else if(in[i]>out[i])w[i][t].c=(in[i]-out[i])/2,maxn+=(in[i]-out[i])/2; } } int main(){ int T; scanf("%d",&T); while(T--){ memset(w,0,sizeof(w)); maxn=0; init(); solve(); if(flag) dinic(); } return 0; }