标签:blog os io for ar div cti log amp
其实就是让你找最少的拐弯次数,dk数组记录到一个点的最少拐弯次数,每次让一个方向上的所有点进队就好了。
注意如果拐弯次数相等还是可以进队的,因为过来的方向可能不一样。
#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_back
typedef 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 = 110;
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
int dk[maxn][maxn];
char mp[maxn][maxn];
int sx,sy,ex,ey,n,m,k;
void bfs() {
queue<int> qx,qy,qk;
qx.push(sx); qy.push(sy);
dk[sx][sy] = 0;
while(!qx.empty()) {
int x = qx.front(), y = qy.front(), nk = dk[x][y];
qx.pop(); qy.pop();
for(int i = 0;i < 4;i++) {
int nx = x + dx[i], ny = y + dy[i];
while(mp[nx][ny] != ‘*‘ && dk[x][y] + 1 <= dk[nx][ny]) {
qx.push(nx); qy.push(ny);
dk[nx][ny] = dk[x][y] + 1;
nx += dx[i]; ny += dy[i];
}
}
}
}
int main() {
int T; scanf("%d",&T);
while(T--) {
memset(dk,0x3f,sizeof(dk));
memset(mp,‘*‘,sizeof(mp));
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++) {
for(int j = 1;j <= m;j++) {
scanf(" %c",&mp[i][j]);
}
}
scanf("%d%d%d%d%d",&k,&sy,&sx,&ey,&ex);
bfs();
if(dk[ex][ey] > k + 1) puts("no");
else puts("yes");
}
return 0;
}
标签:blog os io for ar div cti log amp
原文地址:http://www.cnblogs.com/rolight/p/3931733.html