题目大意:给出一个无向边,非常多询问,问x,y两地之间的最长路最短是多少。
思路:乍一看好像是二分啊。
的确这个题二分能够做。可是时间会慢非常多,有的题直接就T掉(NOIP2013货车运输)。
事实上这个题的模型就是最小瓶颈路模型。
解法就是把无向图变成一个最小生成树,然后两点之间的最长路就是满足题意的答案。
CODE:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 15010 #define INF 0x7f7f7f7f using namespace std; struct Complex{ int x,y,len; bool operator <(const Complex &a)const { return len < a.len; } void Read() { scanf("%d%d%d",&x,&y,&len); } }edge[MAX << 2]; int points,edges,asks; int head[MAX],total; int next[MAX << 1],length[MAX << 1],aim[MAX << 1]; int fa[MAX]; int deep[MAX]; int father[MAX][20],min_length[MAX][20]; void Pretreatment(); int Find(int x); inline void Add(int x,int y,int len); void DFS(int x,int last); void SparseTable(); int GetLCA(int x,int y); int main() { cin >> points >> edges >> asks; Pretreatment(); for(int i = 1;i <= edges; ++i) edge[i].Read(); sort(edge + 1,edge + edges + 1); for(int i = 1;i <= edges; ++i) { int fx = Find(edge[i].x); int fy = Find(edge[i].y); if(fx != fy) { Add(edge[i].x,edge[i].y,edge[i].len); Add(edge[i].y,edge[i].x,edge[i].len); fa[fx] = fy; } } DFS(1,0); SparseTable(); for(int x,y,i = 1;i <= asks; ++i) { scanf("%d%d",&x,&y); printf("%d ",GetLCA(x,y)); } return 0; } void Pretreatment() { for(int i = 1;i <= points; ++i) fa[i] = i; } int Find(int x) { if(fa[x] == x) return x; return fa[x] = Find(fa[x]); } inline void Add(int x,int y,int len) { next[++total] = head[x]; aim[total] = y; length[total] = len; head[x] = total; } void DFS(int x,int last) { deep[x] = deep[last] + 1; for(int i = head[x];i;i = next[i]) { if(aim[i] == last) continue; father[aim[i]][0] = x; min_length[aim[i]][0] = length[i]; DFS(aim[i],x); } } void SparseTable() { for(int j = 1;j < 20; ++j) for(int i = 1;i <= points; ++i) { father[i][j] = father[father[i][j - 1]][j - 1]; min_length[i][j] = max(min_length[i][j - 1],min_length[father[i][j - 1]][j - 1]); } } int GetLCA(int x,int y) { int re = 0; if(deep[x] < deep[y]) swap(x,y); for(int i = 19;i >= 0; --i) if(deep[father[x][i]] >= deep[y]) { re = max(re,min_length[x][i]); x = father[x][i]; } if(x == y) return re; for(int i = 19;i >= 0; --i) if(father[x][i] != father[y][i]) { re = max(re,min_length[x][i]); re = max(re,min_length[y][i]); x = father[x][i]; y = father[y][i]; } re = max(re,min_length[x][0]); re = max(re,min_length[y][0]); return re; }