CH2601 电路维修(算竞进阶习题)

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;

}