【题意】在一个无向图中，一个人要从A点赶往B点，之后再赶往C点，且要求中途不能多次经过同一个点。问是否存在这样的路线。（3 <= N <= 30011, 1 <= M <= 50011） 【思路】很巧的一道题，一般我们都是把源点连接起点，但那样的话就不好控制从A先到B再到C了，所以我们换个思路，以B为源点，A、C为汇点，看最大流是否为2即可~不经过同一个点就直接拆点连一条(i, i', 1)即可，无向图……就连两条反向边吧~~本来想改一下反向流就好的，可是想想那样也把源点汇点连出来的边也变成双向了……没试行不行……

#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 60055;
const int MAXE = 200055;
const int oo = 0x3fffffff;
struct node{int u, v, flow;int opp;int next;
};
struct Dinic{node arc[2*MAXE];int vn, en, head[MAXV];     //vn点个数(包括源点汇点),en边个数int cur[MAXV];              //当前弧int q[MAXV];                //bfs建层次图时的队列int path[2*MAXE], top;        //存dfs当前最短路径的栈int dep[MAXV];              //各节点层次void init(int n){vn = n;en = 0;mem(head, -1);}void insert_flow(int u, int v, int flow){arc[en].u = u;arc[en].v = v;arc[en].flow = flow;arc[en].opp = en + 1;arc[en].next = head[u];head[u] = en ++;arc[en].u = v;arc[en].v = u;arc[en].flow = 0;       //反向弧arc[en].opp = en - 1;arc[en].next = head[v];head[v] = en ++;}bool bfs(int s, int t){mem(dep, -1);int lq = 0, rq = 1;dep[s] = 0;q[lq] = s;while(lq < rq){int u = q[lq ++];if (u == t){return true;}for (int i = head[u]; i != -1; i = arc[i].next){int v = arc[i].v;if (dep[v] == -1 && arc[i].flow > 0){dep[v] = dep[u] + 1;q[rq ++] = v;}}}return false;}int solve(int s, int t){int maxflow = 0;while(bfs(s, t)){int i, j;for (i = 1; i <= vn; i ++)  cur[i] = head[i];for (i = s, top = 0;;){if (i == t){int mink;int minflow = 0x3fffffff;for (int k = 0; k < top; k ++)if (minflow > arc[path[k]].flow){minflow = arc[path[k]].flow;mink = k;}for (int k = 0; k < top; k ++)arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;maxflow += minflow;top = mink;       //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).i = arc[path[top]].u;}for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){int v = arc[j].v;if (arc[j].flow && dep[v] == dep[i] + 1)break;}if (j != -1){path[top ++] = j;i = arc[j].v;}else{if (top == 0)   break;dep[i] = -1;i = arc[path[-- top]].u;}}}return maxflow;}
}dinic;
int main(){int t;scanf("%d",&t);while(t --){int n, m;scanf("%d %d", &n, &m);if (n < 3){puts("NO");continue;}dinic.init(2*n+2);for (int i = 1; i <= n; i ++){dinic.insert_flow(2*i-1, 2*i, 1);}for (int i = 1; i <= m; i ++){int u, v;scanf("%d %d", &u, &v);if(u > n || v > n) continue;dinic.insert_flow(2*u, 2*v-1, 1);dinic.insert_flow(2*v, 2*u-1, 1);}dinic.insert_flow(2*n+1, 2*2, 2);dinic.insert_flow(2*1, 2*n+2, 1);dinic.insert_flow(2*3, 2*n+2, 1);if (dinic.solve(2*n+1, 2*n+2) == 2){puts("YES");}else{puts("NO");}}return 0;
}