題目大意:給出一張混合圖,問是否存在歐拉回路。
思路:成題,直接看題解吧。
CODE:
#include#include #include #include #include #define MAX 510 #define MAXE 5000000 #define INF 0x3f3f3f3f #define S 0 #define T (MAX - 1) using namespace std; #define min(a,b) ((a) < (b) ? (a):(b)) #define max(a,b) ((a) > (b) ? (a):(b)) struct Edge{ int x,y; bool directed; void Read() { int z; scanf(%d%d%d,&x,&y,&z); directed = z; } }edge[MAXE]; struct MaxFlow{ int head[MAX],total; int next[MAXE],aim[MAXE],flow[MAXE]; int deep[MAX]; void Initialize() { total = 1; memset(head,0,sizeof(head)); } void Add(int x,int y,int f) { next[++total] = head[x]; aim[total] = y; flow[total] = f; head[x] = total; } void Insert(int x,int y,int f) { Add(x,y,f); Add(y,x,0); } bool BFS() { static queue q; while(!q.empty()) q.pop(); memset(deep,0,sizeof(deep)); deep[S] = 1; q.push(S); while(!q.empty()) { int x = q.front(); q.pop(); for(int i = head[x]; i; i = next[i]) if(flow[i] && !deep[aim[i]]) { deep[aim[i]] = deep[x] + 1; q.push(aim[i]); if(aim[i] == T) return true; } } return false; } int Dinic(int x,int f) { if(x == T) return f; int temp = f; for(int i = head[x]; i; i = next[i]) if(deep[aim[i]] == deep[x] + 1 && flow[i] && temp) { int away = Dinic(aim[i],min(flow[i],temp)); flow[i] -= away; flow[i^1] += away; temp -= away; } return f - temp; } }solver; int _T; int points,edges; int in[MAX],out[MAX]; int sum; inline bool Euler() { sum = 0; for(int i = 1; i <= points; ++i) { int degree = out[i] - in[i]; if(degree&1) return false; degree >>= 1; if(degree > 0) solver.Insert(S,i,degree),sum += degree; else solver.Insert(i,T,-degree); } return true; } int main() { for(cin >> _T; _T--;) { solver.Initialize(); memset(in,0,sizeof(in)); memset(out,0,sizeof(out)); scanf(%d%d,&points,&edges); for(int i = 1; i <= edges; ++i) { edge[i].Read(); if(edge[i].x == edge[i].y) continue; if(!edge[i].directed) solver.Insert(edge[i].x,edge[i].y,1); ++in[edge[i].y],++out[edge[i].x]; } if(!Euler()) { puts(impossible); continue; } int max_flow = 0; while(solver.BFS()) max_flow += solver.Dinic(S,INF); if(max_flow == sum) puts(possible); else puts(impossible); } return 0; }