题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4612
给一个无向图, 加上一条边后,求桥至少有几个;
那我们加的那条边的两个顶点u,v;一定是u,v之间含有桥的数量最多,然后uv之间的桥都没了,剩下的就是要求的结果;
树的直径的定义刚好就是两个节点之间含有最多的边;
下面是有关树的直径的知识;
这个题目需要手动扩展,不然会爆栈,而且手动扩展的话要用C++提交。
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; #define N 200005 int Head1[N], Head2[N], cnt[3]; int Stack[N], top, dfn[N], low[N], Time, n, m; int nBridge, Bridge[N]; int dist[N], vis[N], Max, index; struct Edge { int v, next; } e1[10*N], e2[10*N]; void Init() { top = nBridge = Time = Max = index = 0; cnt[1] = cnt[0] = 0; memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(Bridge, 0, sizeof(Bridge)); memset(dist, 0, sizeof(dist)); memset(Stack, 0, sizeof(Stack)); memset(Head1, -1, sizeof(Head1)); memset(Head2, -1, sizeof(Head2)); } void Add(Edge e[],int Head[], int u, int v, int k) { e[cnt[k]].v = v; e[cnt[k]].next = Head[u]; Head[u] = cnt[k]++; } void Tarjar(int u, int father) { low[u] = dfn[u] = ++Time; Stack[top++] = u; int v, k=0; for(int i=Head1[u]; i!=-1; i=e1[i].next) { v = e1[i].v; if(v==father && !k)///避免重边; { k++; continue; } if(!dfn[v]) { Tarjar(v, u); low[u] = min(low[u], low[v]); } else low[u] = min(low[u], dfn[v]); } if(low[u] == dfn[u]) { nBridge++;///可以代表缩点后的节点个数; while(1) { v = Stack[--top]; Bridge[v] = nBridge;///缩点; if(u==v) break; } } } void bfs(int s) { queue<int>Q; int p, q; memset(vis, 0, sizeof(vis)); vis[s] = 1; dist[s] = 0; Q.push(s); while(!Q.empty()) { p = Q.front(); Q.pop(); for(int i=Head2[p]; i!=-1; i=e2[i].next) { q = e2[i].v; if(!vis[q]) { vis[q] = 1; dist[q] = dist[p] + 1; Q.push(q); if(Max<dist[q]) { Max = dist[q]; index = q; } } } } } int main() { int u, v; while(scanf("%d%d", &n, &m), m + n)///输入时由于m n弄反了,TLE的我想哭; { Init(); for(int i=1; i<=m; i++) { scanf("%d%d", &u, &v); Add(e1, Head1, u, v, 0); Add(e1, Head1, v, u, 0);///原来的树; } Tarjar(1, 0); for(int i=1; i<=n; i++) { for(int j=Head1[i]; j!=-1; j=e1[j].next) { int u = Bridge[i]; int v = Bridge[e1[j].v]; if(u != v ) { Add(e2, Head2, u, v, 1); Add(e2, Head2, v, u, 1);///缩点后的树; } } } bfs(1); bfs(index);///求树的直径的过程; printf("%d\n", nBridge-1-Max);///缩点后形成的树每条边都是桥;所以总桥的个数为节点数-1; } return 0; }
时间: 2024-11-05 13:45:52