POJ 637 Sightseeing tour 混合欧拉回路 最大流

Sightseeing tour

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9114 Accepted: 3841

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

45 82 1 01 3 04 1 11 5 05 4 13 4 04 2 12 2 04 41 2 12 3 03 4 01 4 13 31 2 02 3 03 2 03 41 2 02 3 11 2 03 2 0

Sample Output

possibleimpossibleimpossiblepossible

题意：给一个既有有向边又有无向边的图，问这个图是否存在欧拉回路。

       欧拉回路就是从一个点出发，经过每条边一次并回到起点的路径。这个只问是否存在，不用找路径。对于有向图，要求每个点的入度==出度就有欧拉回路，关于欧拉回路：点击打开链接

       做法：要使有欧拉回路，就要使每个点的入度==出度，方法是网络流，首先是将无向边任意定向，计算每个点的入度和出度，如果有某个点的出入度之差为奇数，就肯定不存在欧拉回路，为偶数可以改变与这个点连接的无向边的方向来改变出入度（有向边反向点的出度+1，入度-1或者出度-1，入度+1）。这是一个判断条件。

       然后就是关于网络流建模，建立源点s，汇点t。有向边方向已定，不能改变点的出入度，计算了关于有向边的点的出入度后，删掉。我们只关心无向边的方向，所以只在图中保存随意定向的无向边，流量为1。对于出度>入度的点i，它需要入度，建边(s,i,(out[i]-in[i])/2)，出度<入度，需要出度，建边(i,t,(in[i]-out[i])/2)。然后跑一次最大流，如果能满流就是存在欧拉回路的，否则不存在。

       关于正确性，每一次可行流必然是由s出发，经过一个出>入的点，最后经过一个入>出的点到t，将这个可行流上的边都反向，是不是对于出>入的点出度-1，入度+1，入>出的点出度+1，入度-1，这里不考虑s和t连的边对点的出入度影响，因为这只是为了求解最大流新建的边，不是原图中的边，对原图中的点的出入度没有影响。

#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>#include <queue>using namespace std;const int N = 300+10;const int INF = 1e8;struct node{    int v,flow,next;    node(){}    node(int v,int flow,int next):        v(v),flow(flow),next(next){}}E[N*100];int n,m,top,s,t;int head[N];  ///邻接表头结点int in[N];    ///入度int out[N];   ///出度int d[N];     ///dinic的disbool vis[N];  ///dinic用void Init(){    top = 0;    for(int i = 0;i < N;i++){        head[i] = -1;        in[i] = out[i] = 0;    }}void add(int u,int v,int flow){    E[top] = node(v,flow,head[u]);    head[u] = top++;    E[top] = node(u,0,head[v]);    head[v] = top++;}bool bfs(){    memset(vis,false,sizeof vis);    memset(d,0,sizeof d);    queue<int>q;    vis[s] = true;    q.push(s);    while(!q.empty()){        int u = q.front();        q.pop();        for(int i = head[u];i != -1;i = E[i].next){            int v = E[i].v;            if(vis[v] || E[i].flow == 0) continue;            d[v] = d[u]+1;            vis[v] = true;            q.push(v);        }    }    return vis[t];}int dfs(int u,int a){    if(u == t || a == 0){        return a;    }    int flow = 0;    for(int i = head[u];i != -1;i = E[i].next){        int v = E[i].v;        if(d[v] != d[u]+1 || E[i].flow == 0) continue;        int f = dfs(v,min(a,E[i].flow));        if(f > 0){            a -= f;            E[i].flow -= f;            E[i^1].flow += f;            flow += f;        }    }    return flow;}int dinic()  ///小红书最大流模板{    int flow = 0;    while(bfs()){        flow += dfs(s,INF);    }    return flow;}int main(void){    int T;    scanf("%d",&T);    while(T--){        Init();        s = 210,t = 211;        scanf("%d%d",&n,&m);        for(int i = 1;i <= m;i++){            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            out[u]++;            in[v]++;            if(w == 0){     ///只建无向向边的图，定向u->v                add(u,v,1);            }        }        bool flag = false;        for(int i = 1;i <= n;i++){            if(abs(in[i]-out[i]%2 == 1))  ///出入度差为奇数                flag = true;            else{                if(in[i] > out[i])                    add(i,t,(in[i]-out[i])/2);                if((in[i] < out[i]))                    add(s,i,(out[i]-in[i])/2);            }        }        if(flag){            printf("impossible\n");            continue;        }        int tot = 0;  ///满流总量        for(int i = head[s];i != -1;i = E[i].next){            tot += E[i].flow;        }        int ans = dinic();        if(ans == tot)            printf("possible\n");        else            printf("impossible\n");    }    return 0;}