CH2601 电路维修(算竞进阶习题) た 入场券 2022-01-07 05:11 151阅读 0赞 # 01边bfs # 这题很容易想到的就是根据符号的情况建图,把每个点方格的对角线看成图的节点,有线相连就是边权就是0,没有就是1 然后跑最短路,但是最短路用的优先队列维护是有logn的代价的 这题还有一个更快的方法,就是双端队列。。0边放队头,1边放队尾,然后虽然每个点会入队多次,但是我们只要取第一次出队(最短的情况)就行了 时间复杂度为O(r\*c) #include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; inline int lowbit(int x){ return x & (-x); } inline int read(){ int X = 0, w = 0; char ch = 0; while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); } while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar(); return w ? -X : X; } inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; } inline int lcm(int a, int b){ return a / gcd(a, b) * b; } template<typename T> inline T max(T x, T y, T z){ return max(max(x, y), z); } template<typename T> inline T min(T x, T y, T z){ return min(min(x, y), z); } template<typename A, typename B, typename C> inline A fpow(A x, B p, C lyd){ A ans = 1; for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd; return ans; } char g[510][510]; int r, c, cnt, head[510*510], d[510*510], ed; bool vis[510*510]; struct Edge{ int v, next, dis; }edge[500000<<1]; void addEdge(int a, int b, int w){ edge[cnt].v = b, edge[cnt].dis = w, edge[cnt].next = head[a]; head[a] = cnt ++; } int bfs(){ memset(d, -1, sizeof d); memset(vis, 0, sizeof vis); deque<pair<int, int>> q; d[1] = 0; q.push_front(make_pair(1, 0)); while(!q.empty()){ int s = q.front().first, d = q.front().second; q.pop_front(); if(s == ed) return d; if(vis[s]) continue; vis[s] = true; for(int i = head[s]; i != -1; i = edge[i].next){ int u = edge[i].v; if(!vis[u]){ if(edge[i].dis == 1) q.push_back(make_pair(u, d + 1)); else q.push_front(make_pair(u, d)); } } } return -1; } int main(){ int _; scanf("%d", &_); for(; _; _ --){ memset(head, -1, sizeof head); cnt = 0; scanf("%d%d", &r, &c); for(int i = 1; i <= r; i ++) scanf("%s", g[i] + 1); for(int i = 1; i <= r; i ++){ for(int j = 1; j <= c; j ++){ int a = (i - 1) * (c + 1) + j, b = a + 1; int d = a + c + 2, c = d - 1; if(g[i][j] == '\\'){ addEdge(a, d, 0), addEdge(d, a, 0); addEdge(b, c, 1), addEdge(c, b, 1); } else{ addEdge(a, d, 1), addEdge(d, a, 1); addEdge(b, c, 0), addEdge(c, b, 0); } } } //for(int i = 0; i < cnt; i ++) cout << edge[i].v << " " << edge[i].dis << endl; ed = (r + 1) * (c + 1); int ans = bfs(); printf(ans == -1 ? "NO SOLUTION\n" : "%d\n", ans); } return 0; } 转载于:https://www.cnblogs.com/onionQAQ/p/10566993.html
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