newcode wyh的吃鸡（优势队列+BFS）题解

思路：

要用优势队列，因为有的+2，有的+1，所以队列中的步长是不单调的，所以找到一个答案但不一定最小，所以用优势队列把小的放在队首。

要记录状态，所以开了三维，题目和昨天做的那道小明差不多

vis开的int型赋值bool型WA了半天

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int N = 100+5;
char mp[N][N];
int vis[3][N][N],k,x,y,n,to[4][2] = {0,1,0,-1,1,0,-1,0};
struct node{
    int x,y;
    int step,speed;
    bool operator < (const node &a) const{
        return step > a.step;
    }
};
void BFS(){
    priority_queue<node> q;
    while(!q.empty()) q.pop();
    node a,b;
    a.x = x,a.y = y,a.speed = 2,a.step = 0;
    vis[a.speed][a.x][a.y] = 1;
    q.push(a);
    while(!q.empty()){
        a = q.top();
        q.pop();
        if(mp[a.x][a.y] == 'X' && a.step <= k){
            printf("YES
%d
",a.step);
            return;
        }
        for(int i = 0;i < 4;i++){
            b.x = a.x + to[i][0];
            b.y = a.y + to[i][1];
            b.speed = a.speed;
            if(b.x < 1 || b.y < 1 || b.x > n || b.y > n) continue;
            if(mp[b.x][b.y] == 'O') continue;
            if(vis[b.speed][b.x][b.y]) continue;
            vis[b.speed][b.x][b.y] = 1;
            b.step = a.step + b.speed;
            if(mp[b.x][b.y] == 'C') b.speed = 1;
            if(b.step <= k) q.push(b);
        }
    }
    printf("NO
");
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&k);
        for(int i = 1;i <= n;i++){
            scanf("%s",mp[i] + 1);
            for(int j = 1;j <= n;j++){
                if(mp[i][j] == 'S') x = i,y = j;
            }
        }
        memset(vis,0,sizeof(vis));
        BFS();
    }
    return 0;
}