傳送門:洛谷P2243

題目描述

分析

初步判斷,是道最短路的題.
　　首先就是建圖了,對於電路板上的每一對角線,令與讀入方向相同的路徑邊權為0,方向相反的邊權為1.
　　由於只能走斜線,根據網格圖的性質可知橫縱座標之和為奇數的點是到不了的,預設起點為 $(0,0)$

程式碼

#include <cstdio>
#include <cstdlib>
#include
 
 <cstring>
#include <deque>

#define IL inline

using namespace std;

IL int read()
{
    char c = getchar();
    int sum = 0 ,k = 1;
    for(;'0' > c || c > '9'; c = getchar())
        if(c == '-') k = -1;
    for(;'0' <= c && c <= '9'; c = getchar()) sum = sum * 10 + c -
 
 '0';
    return sum * k;
}

int to[502005], nxt[502005], val[502005];
int last[251005];
int cnt;
IL void add(int u, int v, int w)
{
    //printf("%d %d %d\n", u, v, w);
    to[++cnt] = v; nxt[cnt] = last[u]; val[cnt] = w; last[u] = cnt;
    to[++cnt] = u; nxt[cnt] = last[v]; val[cnt] = w; last[v] = cnt;
}

int S, T;
int n, m;

int dis[251005];

IL int turn(int x, int y)
{	
    return x * (m + 1) + y + 1;
}

IL void bfs(int u)
{
    memset(dis, -1, sizeof(dis));
    dis[u] = 0;
    deque<int>Q;
    Q.push_back(u);
    if(S == T) { printf("0\n"); return ; }
    for(; !Q.empty();)
    {
        u = Q.front(); Q.pop_front();
        for(int i = last[u], v; (v = to[i]); i = nxt[i])
        if(dis[v] == -1 || dis[u] + val[i] < dis[v])
        {	
            dis[v] = dis[u] + val[i];
            //printf("%d %d %d %d\n", u, v, dis[u], dis[v]);
            if(v == T) { printf("%d\n", dis[v]); return ; }
            if(!val[i]) Q.push_front(v); else Q.push_back(v);
        }
    }
}

int main()
{
    for(int k = read(); k; --k)
    {
        cnt = 0;
        memset(last, 0, sizeof(last));
    	n = read(); m = read();
    	char c;
    	for(int i = 0; i < n; ++i)
   		for(int j = 0; j < m; ++j)
    	{
    		scanf(" %c", &c);
    		if((i + j) & 1) add(turn(i, j + 1), turn(i + 1, j), c == '\\');
    		else add(turn(i, j), turn(i + 1, j + 1), c == '/');
    	}
    	if((n + m) & 1) { printf("NO SOLUTION\n"); continue; }
    	S = 1; T = turn(n, m);
    	bfs(S);
    }
    return 0;
}