【

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;}