CH2601 电路维修(算竞进阶习题)

た入场券 2022-01-07 05:11 202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