題意：網(wǎng)格圖，老鼠吃奶酪，吃完奶酪體力值會(huì)增加，只能吃硬度不大于體力值的，問(wèn)最小步數(shù)。

思路：按硬度從小到大的吃起，依次求最短路。

我用曼哈頓距離估價(jià)的A*，和普通bfs的time沒(méi)區(qū)別啊，還把優(yōu)先級(jí)那里寫(xiě)錯(cuò)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 LOCALfreopen("in.txt","r",stdin); #endifH = 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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/jerryRey/p/4889607.html

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