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HDU 2579 Dating with girls(2) BFS 余数判重

对于石头的处理就按照每个位置的时间取k的余数判一下重复就好,其他随意写

#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 105;const int maxk = 10;const int dx[] = {0,0,1,-1};const int dy[] = {1,-1,0,0};int d[maxn][maxn][maxk];int n,m,k,sx,sy,ex,ey;char mp[maxn][maxn];void bfs() {    queue<int> qx,qy,qk;    qx.push(sx); qy.push(sy); qk.push(0);    d[sx][sy][0] = 0;    int x,y,nowk,nx,ny,nk;    while(!qx.empty()) {        x = qx.front(); y = qy.front(); nowk = qk.front();        qx.pop(); qy.pop(); qk.pop();        int nowt = d[x][y][nowk];        for(int i = 0;i < 4;i++) {            nx = x + dx[i]; ny = y + dy[i]; nk = (nowk + 1) % k;            int &nt = d[nx][ny][nk];            if((nk == 0 || mp[nx][ny] != ‘#‘) && nt > nowt + 1) {                if(nx < 1 || nx > n || ny < 1 || ny > m) continue;                nt = nowt + 1;                qx.push(nx); qy.push(ny); qk.push(nk);            }        }    }}int main() {    int T; scanf("%d",&T);    while(T--) {        memset(mp,‘#‘,sizeof(mp));        memset(d,0x3f,sizeof(d));        int inf = d[0][0][0];        scanf("%d%d%d",&n,&m,&k);        for(int i = 1;i <= n;i++) {            for(int j = 1;j <= m;j++) {                scanf(" %c",&mp[i][j]);                if(mp[i][j] == ‘Y‘) {                    sx = i; sy = j;                }                if(mp[i][j] == ‘G‘) {                    ex = i; ey = j;                }            }        }        bfs();        int ans = inf;        for(int i = 0;i < k;i++) ans = min(ans,d[ex][ey][i]);        if(ans < inf) printf("%d\n",ans);        else puts("Please give me another chance!");    }    return 0;}

  

HDU 2579 Dating with girls(2) BFS 余数判重