poj 1637 Sightseeing tour（混合欧拉）

Sightseeing tour

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6937		Accepted: 2877

DescriptionThe 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 exactlyonce. 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 tourunder these constraints.InputOn 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 numberof 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 streetis 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.OutputFor 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

题意：给出一个混合图，有无向边和有向边，问是否存在欧拉回路。

思路：首先对于每条无向边，随便定向，把它变为有向边。对于每条边（包括原来有向的和后来变有向的），统计出度和入度，若存在一个点|出度-入度|为奇数，那么就不存在欧拉回路。如果不存在这样的点，就要用网络流来判断。

构图如下：

对于某个点，若出度>入度，则从源点连一条边到该点，容量为（出度-入度）/2； 若入度>出度，则从该点连一条边到汇点，容量为（入度-出度）/2。

然后，对于每条原来是无向的边，根据前面的定向，连一条边，容量为1。

最后，判断是否满流。若满流，则存在欧拉回路。

AC代码：

#include <iostream>#include <cmath>#include <cstdlib>#include <cstring>#include <cstdio>#include <queue>#include <ctime>#include <algorithm>#define ll __int64using namespace std;const int INF = 1000000000;const int maxn = 205;struct Edge{int u, v, cap, flow, next;} et[maxn * maxn];int low[maxn], cnt[maxn], cur[maxn], dis[maxn], pre[maxn], eh[maxn];int in[maxn], out[maxn];int n, m, s, t, num;void init(){memset(eh, -1, sizeof(eh));num = 0;}void add(int u, int v, int cap, int flow){Edge e = {u, v, cap, flow, eh[u]};et[num] = e;eh[u] = num++;}void addedge(int u, int v, int cap){add(u, v, cap, 0);add(v, u, 0, 0);}int isap(int s, int t, int nv){int u, v, now, flow = 0;memset(cnt, 0, sizeof(cnt));memset(dis, 0, sizeof(dis));memset(low, 0, sizeof(low));for(u = 0; u <= nv; u++) cur[u] = eh[u];low[s] = INF, cnt[0] = nv, u = s;while(dis[s] < nv){for(now = cur[u]; now != -1; now = et[now].next)if(et[now].cap - et[now].flow && dis[u] == dis[v = et[now].v] + 1) break;if(now != -1){cur[u] = pre[v] = now;low[v] = min(et[now].cap - et[now].flow, low[u]);u = v;if(u == t){for(; u != s; u = et[pre[u]].u){et[pre[u]].flow += low[t];et[pre[u]^1].flow -= low[t];}flow += low[t];low[s] = INF;}}else{if(--cnt[dis[u]] == 0) break;dis[u] = nv, cur[u] = eh[u];for(now = eh[u]; now != -1; now = et[now].next)if(et[now].cap - et[now].flow && dis[u] > dis[et[now].v] + 1)dis[u] = dis[et[now].v] + 1;cnt[dis[u]]++;if(u != s) u =et[pre[u]].u;}}return flow;}int main(){int tt;int a, b, c;scanf("%d", &tt);while(tt--){scanf("%d%d", &n, &m);init();s = 0;t = n + 1;memset(in, 0, sizeof(in));memset(out, 0, sizeof(out));while(m--){scanf("%d%d%d", &a, &b, &c);out[a]++;in[b]++;if(c != 1) addedge(a, b, 1);}bool possible = true;int sum = 0;for(int i = 1; i <= n; i++){if(abs(out[i] - in[i]) & 1){possible = false;break;}if(out[i] > in[i]){addedge(s, i, (out[i] - in[i]) / 2);sum += (out[i] - in[i]) / 2;}else if(in[i] > out[i]) addedge(i, t, (in[i] - out[i]) / 2);}if(!possible){printf("impossible\n");continue;}if(isap(s, t, t + 1) == sum) printf("possible\n");else printf("impossible\n");}return 0;}