UVA 11165 - Galactic Travel
题意:给定一些不能走的边,要求出从s到t的最短路
思路:由于点数多,直接广搜会超时,所以加上优化,已经找过的点就不在重复找了,这点可以利用并查集进行优化,然后对于每个点的每个不能走的区间,可以先排序,然后利用twopointer的性质,每次可以从上次找到的位置往后找即可
代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; const int N = 100005; int t, n, m, parent[N], vis[N]; int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]); } struct Ban { int v1, v2; Ban(int v1, int v2) { this->v1 = v1; this->v2 = v2; } }; bool cmp(Ban a, Ban b) { return a.v1 < b.v1; } vector<Ban> b[N]; const int INF = 0x3f3f3f3f; int bfs(int s, int t) { queue<int> Q; Q.push(s); vis[s] = 0; while (!Q.empty()) { int u = Q.front(); Q.pop(); if (u == t) return vis[u]; int v = find(0); int s = 0; while (v < n) { if (v == u) v = find(v + 1); else { int flag = 1; for (int i = s; i < b[u].size(); i++) { if (b[u][i].v1 <= v && b[u][i].v2 >= v) { flag = 0; v = find(b[u][i].v2 + 1); s = i + 1; break; } } if (flag) { vis[v] = vis[u] + 1; Q.push(v); parent[v] = find(v + 1); v = find(v); } } } } return -1; } int main() { int cas = 0; scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); for (int i = 0; i <= n; i++) { parent[i] = i; b[i].clear(); } int u, v1, v2; while (m--) { scanf("%d%d-%d", &u, &v1, &v2); if (v1 > v2) swap(v1, v2); b[u].push_back(Ban(v1, v2)); } for (int i = 0; i < n; i++) sort(b[i].begin(), b[i].end(), cmp); int s, t; scanf("%d%d", &s, &t); printf("Case #%d: ", ++cas); int tmp = bfs(s, t); if (tmp == -1) printf("Impossible\n"); else printf("%d\n", tmp); } return 0; }
时间: 2024-10-11 00:31:37