UVa 10735

题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=1676&mosmsg=Submission+received+with+ID+26560367

先将混合图中的无向图定向，然后统计每个点的入度和出度

有向图存在欧拉回路的充要条件是：所有点的入度等于出度

则出度大于入度的点需要减少出度，出度小于入度的点需要增加出度，改变出度的方法是将边反向，

将边反向的操作相当于将出度“运送”到下一个点，所以问题相当于将多余的出度运送到缺少的出度

将源点连接所有剩余出度的点，容量为 ((out[i] - in[i])/2), 缺少出度的点连接汇点, 容量为 ((in[i] - out[i])/2)

所有定向后的无向边相连，容量为 (1)，如果有流量流过表示将该边反向

最后建出新图后求出欧拉回路即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1010;
const int INF = 1000000007;

int T, n, m;
int u[maxn], v[maxn], dir[maxn], rev[maxn];
int in[maxn], out[maxn];

int h[maxn], cnt = 1;
struct E{
	int to, cap, next;
	int id;
}e[maxn * 100];
void add(int u, int v, int c, int id){
	e[++cnt].to = v;
	e[cnt].cap = c;
	e[cnt].id = id; // 边的编号 
	e[cnt].next = h[u];
	h[u] = cnt;
}

int s,t;
int vis[maxn], d[maxn], cur[maxn];

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int u = Q.front(); Q.pop();
      for(int i = h[u]; i != -1 ; i = e[i].next) {
        if(!vis[e[i].to] && e[i].cap) {
          vis[e[i].to] = 1;
          d[e[i].to] = d[u] + 1;
          Q.push(e[i].to);
        }
      }
    }
    return vis[t];
  }

  int DFS(int x, int a) {
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1 ; i = e[i].next) {
      if(d[x] + 1 == d[e[i].to] && (f = DFS(e[i].to, min(a, e[i].cap))) > 0) {
        e[i].cap -= f;
        e[i^1].cap += f;
        flow += f;
        a -= f;
        if(a == 0) break;
      }
    }
    return flow;
  }

  int Maxflow() {
    int flow = 0;
    while(BFS()) {
	  memcpy(cur, h, sizeof(h));
      flow += DFS(s, INF);
    }
    return flow;
  }

vector<int> path;
int vi[maxn * 100];
void euler(int u){
	for(int i = h[u] ; i != -1 ; i = e[i].next){
		int v = e[i].to;
		if(!vi[e[i].id]){
			vi[e[i].id] = 1;
			euler(e[i].to);
			path.push_back(e[i].to);
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	scanf("%d", &T);
	int fb = 0;
	while(T--){
		if(fb) printf("
");
		fb = 1;
		
		int flag = 0;
		memset(h, -1, sizeof(h)); cnt = 1;
		memset(in, 0, sizeof(in));
		memset(out, 0, sizeof(out));
		memset(vi, 0, sizeof(vi));
		memset(rev, 0, sizeof(rev));
		path.clear();
		
		scanf("%d%d", &n, &m);
		
		char ty[10]; 
		for(int i = 1 ; i <= m ; ++i){
			scanf("%d%d", &u[i], &v[i]);
			++in[v[i]]; ++out[u[i]];
			
			scanf("%s", ty);
			dir[i] = (ty[0] == 'U' ? 0 : 1);
		}
		
		s = n + 1, t = n + 2;
		int tot = 0;
		for(int i = 1 ; i <= n ; ++i){
			if(out[i] - in[i] == 0) continue;
			if((out[i] - in[i]) & 1) {
				flag = 1;
				printf("No euler circuit exist
");
				break;
			} else{
				if(out[i] - in[i] > 0){ // 源点向提供出边的点连边 
					add(s, i, (out[i] - in[i]) / 2, -1);
					add(i, s, 0, -1);
					tot += (out[i] - in[i]) / 2;
				} else{ // 需要出边的点向汇点连边 
					add(i, t, (in[i] - out[i]) / 2, -1);
					add(t, i, 0, -1);
				}
			}
		}
		
		if(flag) continue;
		
		for(int i = 1 ; i <= m ; ++i){
			if(!dir[i]){
				add(u[i], v[i], 1, i);
				add(v[i], u[i], 0, -1);				
			}
		}
		
		int flow = Maxflow();
		
		if(flow == tot){
			for(int i = 2 ; i <= cnt ; ++i){
				if(e[i].id != -1 && !dir[e[i].id]){
					if(e[i].cap == 0) rev[e[i].id] = 1;
				}
			}
			
			memset(h, -1, sizeof(h)); cnt = 1;
			for(int i = 1 ; i <= m ; ++i){
				if(dir[i]){
					add(u[i], v[i], 0, i);
				} else{
					if(rev[i]) add(v[i], u[i], 0, i);
					else add(u[i], v[i], 0, i);
				}
			}
			
			euler(1);
			
			printf("1 ");
			for(int i = path.size() - 1 ; i >= 0 ; --i){
				printf("%d ", path[i]);
			}
			printf("
");
		} else{
			printf("No euler circuit exist
");
		}
	}
	return 0;
}