SPOJ 962 Intergalactic Map (从A到B再到C的路线)
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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……
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#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; }