SGU 536 Berland Chess(状态压缩 + bfs)
在一个n*m的棋盘上,你有一个white king,然后还有一些(<15)个黑子,每个黑子有一定的攻击范围并且他们不会移动。求吃掉所有黑子的最少步数。黑子的个数很少,所以用状态压缩来表示棋盘上还剩余哪些黑子。在bfs前先要初始化所有黑子状态下的受攻击的点,这个很恶心。。。初始化完了基本就是无脑搜了吧?
#include<algorithm> #include<iostream> #include<cstring> #include<fstream> #include<sstream> #include<cstdlib> #include<vector> #include<string> #include<cstdio> #include<bitset> #include<queue> #include<stack> #include<cmath> #include<map> #include<set> #define FF(i, a, b) for(int i=a; i<b; i++) #define FD(i, a, b) for(int i=a; i>=b; i--) #define REP(i, n) for(int i=0; i<n; i++) #define CLR(a, b) memset(a, b, sizeof(a)) #define debug puts("**debug**") #define LL long long #define PB push_back #define MP make_pair using namespace std; int dwx[8] = {0, 0, 1, -1, 1, 1, -1, -1}; int dwy[8] = {1, -1, 0, 0, 1, -1, 1, -1}; int dkx[8] = {1, 2, 2, 1, -1, -2, -2, -1}; int dky[8] = {2, 1, -1, -2, -2, -1, 1, 2}; const int SET = (1<<15); const int maxn = 20; //can[sta][i][j] = 1 : 当前剩余sta中的黑子 点<i, j>不可走 //vis[sta][i][j] = 1 :当前剩余sta中的黑子时,点<i, j>已走过 bool can[SET][maxn][maxn], vis[SET][maxn][maxn]; char g[maxn][maxn]; //id[x][y] : 点<x, y>的黑子对应的标号 int n, m, sx, sy, tot, id[maxn][maxn]; inline bool outside(int x, int y) { return x<0 || x>=n || y<0 || y>=m; } //当前剩余sta中的黑子 点x y是否在sta中 inline bool has(int sta, int x, int y) { return isalpha(g[x][y]) && (sta & (1<<id[x][y])); } struct Node { int x, y, st, steps; Node(){} Node(int a, int b, int c, int d):x(a), y(b), st(c), steps(d){} }; void bfs() { queue<Node> q; q.push(Node(sx, sy, tot, 0)); vis[tot][sx][sy] = 1; while(!q.empty()) { Node T = q.front(); q.pop(); if(T.st == 0) { printf("%d\n", T.steps); return ; } REP(i, 8) { int tx = T.x + dwx[i], ty = T.y + dwy[i]; if(outside(tx, ty)) continue; //如果当前状态 包含点<tx, ty>的黑子 if(has(T.st, tx, ty)) { int sta = T.st&(tot-(1<<id[tx][ty])); if(!vis[sta][tx][ty] && !can[sta][tx][ty]) //如果可以吃掉它 { q.push(Node(tx, ty, sta, T.steps+1)); vis[sta][tx][ty] = 1; } } else if(!can[T.st][tx][ty]) { int sta = T.st; if(!vis[sta][tx][ty]) { q.push(Node(tx, ty, sta, T.steps+1)); vis[sta][tx][ty] = 1; } } } } puts("-1"); return ; } //初始化所有状态下 会被攻击的点 void init() { FF(s, 1, tot+1) { //如果s状态下的黑子中 包含g[i][j] 那么更新g[i][j]的攻击范围 REP(i, n) REP(j, m) if(isalpha(g[i][j]) && (s & (1<<id[i][j]))) { if(g[i][j] == 'K') { REP(k, 8) { int tx = i + dkx[k], ty = j + dky[k]; if(outside(tx, ty) || has(s, tx, ty)) continue; can[s][tx][ty] = 1; } } else if(g[i][j] == 'B') { int tx = i, ty = j; while(true) { tx++, ty++; if(outside(tx, ty) || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { tx--, ty--; if(outside(tx, ty) || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { tx++, ty--; if(outside(tx, ty) || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { tx--, ty++; if(outside(tx, ty) || has(s, tx, ty)) break; can[s][tx][ty] = 1; } } else { int tx = i, ty = j; while(true) { tx++; if(tx >= n || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { tx--; if(tx < 0 || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { ty++; if(ty>=m || has(s, tx, ty)) break; can[s][tx][ty] = 1; } tx = i, ty = j; while(true) { ty--; if(ty<0 || has(s, tx, ty)) break; can[s][tx][ty] = 1; } } } } } int main() { scanf("%d%d", &n, &m); tot = 0; REP(i, n) { scanf("%s", g[i]); REP(j, m) { if(g[i][j] == '*') sx = i, sy = j; else if(isalpha(g[i][j])) id[i][j] = tot++; } } if(tot == 0) { puts("0"); return 0; } tot = (1<<tot)-1; init(); bfs(); return 0; }
补充:软件开发 , C++ ,