AOJ 558 Cheese(bfs)

题意：网格图，老鼠吃奶酪，吃完奶酪体力值会增加，只能吃硬度不大于体力值的，问最小步数。

思路：按硬度从小到大的吃起，依次求最短路。

我用曼哈顿距离估价的A*，和普通bfs的time没区别啊，还把优先级那里写错了。。。

#include<bits/stdc++.h>
using namespace std;

#define PS push
#define PB push_back
#define MP make_pair
#define fi first
#define se second
const int INF = 0x3f3f3f3f;

typedef long long ll;

inline int read()
{
int ret; char c; while(c = getchar(),c<'0'||c>'9');
ret = c-'0';
while(c = getchar(),c>='0'&&c<='9') ret = ret*10 + c-'0';
return ret;
}

const int SZ = 1e3+5;

char g[SZ][SZ];
int H,W,N;
int vis[SZ][SZ],clk;
struct Node
{
int x,y,f,h;
bool operator <(const Node&th) const {
return f > th.f || ( f == th.f && h < th.h);//
}
}pos[10];

int tar;
inline int MHT(Node &o)
{
return (abs(pos[tar].x-o.x) + abs(pos[tar].y-o.y));
}

void GetPos()
{
REP0(i,H){
REP0(j,W){
char c = g[i][j];
if(c == 'S'){
pos[0] = {i,j};
}else if('1'<= c && c <='9' ){
pos[c-'0'] = {i,j};
}
}
}
}

const int dx[] = {0,1,0,-1};
const int dy[] = {1,0,-1,0};

inline bool valid(int x,int y)
{
return x>=0&&x<H&&y>=0&&y<W&&g[x][y]!='X'&&vis[x][y] != clk;
}

int astar_bfs(int st)
{
priority_queue<Node> q;
Node u;
u.x = pos[st].x;
u.y = pos[st].y;
u.h = u.f = MHT(u);
q.push(u);
clk++;
while(q.size()){
u = q.top(); q.pop();
if(u.x == pos[tar].x && u.y == pos[tar].y ) return u.f-u.h;
REP0(k,4){
Node v;
v.x = u.x + dx[k];
v.y = u.y + dy[k];
if(!valid(v.x,v.y)) continue;
vis[v.x][v.y] = clk;
v.h = MHT(v);
v.f = u.f-u.h+1+v.h;
q.push(v);
}
}
return -1;
}

//#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
H = read(); W = read(); N = read();
for(int i = 0; i < H; i++){
scanf("%s",g[i]);
}
GetPos();
int ans = 0;
REP1(i,N){
tar = i;
ans += astar_bfs(i-1);
}
printf("%d\n",ans);
return 0;
}