标签：inline signed stdout 一个 bug gcd for comm tin
题意：中文题。
析：是一个简单的搜索，BFS 和 DFS都可行， 主要是这个题有一个坑点，那就是如果有一层是#，另一个层是#或者*，都是过不去的，就可以直接路过，
剩下的就是一个简单的搜索，只不过是两层而已，可能一个就是在#必须传送，这个题目已经说的很清楚了。
代码如下：
BFS：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
char s[2][maxn][maxn];
int vis[2][maxn][maxn];
int t;
struct Node{
int pos, x, y;
Node() { }
Node(int p, int xx, int yy) : pos(p), x(xx), y(yy) { }
bool operator == (const Node &p) const{
return pos == p.pos && x == p.x && y == p.y;
}
};
bool bfs(){
memset(vis, -1, sizeof vis);
Node goal;
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
if(s[0][i][j] == ‘P‘) goal = Node(0, i, j);
else if(s[1][i][j] == ‘P‘) goal = Node(1, i, j);
queue<Node> q;
vis[0][0][0] = 0;
q.push(Node(0, 0, 0));
while(!q.empty()){
Node u = q.front(); q.pop();
if(vis[u.pos][u.x][u.y] > t) return false;
if(u == goal) return true;
for(int i = 0; i < 4; ++i){
int x = u.x + dr[i];
int y = u.y + dc[i];
if(!is_in(x, y) || vis[u.pos][x][y] != -1 || s[u.pos][x][y] == ‘*‘) continue;
vis[u.pos][x][y] = vis[u.pos][u.x][u.y] + 1;
if(s[u.pos][x][y] == ‘#‘){
vis[u.pos^1][x][y] = vis[u.pos][x][y];
if(s[u.pos^1][x][y] == ‘#‘ || s[u.pos^1][x][y] == ‘*‘) continue;
q.push(Node(u.pos^1, x, y));
continue;
}
q.push(Node(u.pos, x, y));
}
}
return false;
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d %d", &n, &m, &t);
for(int i = 0; i < n; ++i) scanf("%s", s[0]+i);
for(int i = 0; i < n; ++i) scanf("%s", s[1]+i);
bool ans = bfs();
printf("%s\n", ans ? "YES" : "NO");
}
return 0;
}
DFS：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
char s[2][maxn][maxn];
bool vis[2][maxn][maxn];
int t;
bool dfs(int pos, int r, int c, int d){
if(d > t) return false;
if(s[pos][r][c] == ‘P‘) return true;
for(int i = 0; i < 4; ++i){
int x = r + dr[i];
int y = c + dc[i];
if(!is_in(x, y) || vis[pos][x][y] || s[pos][x][y] == ‘*‘) continue;
vis[pos][x][y] = 1;
if(s[pos][x][y] == ‘#‘){
vis[pos^1][x][y] = 1;
if(dfs(pos^1, x, y, d+1)) return true;
vis[pos^1][x][y] = 0;
}
else if(dfs(pos, x, y, d+1)) return true;
vis[pos][x][y] = 0;
}
return false;
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d %d", &n, &m, &t);
for(int i = 0; i < n; ++i) scanf("%s", s[0]+i);
for(int i = 0; i < n; ++i) scanf("%s", s[1]+i);
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
if(s[0][i][j] == ‘#‘ && (s[1][i][j] == ‘*‘ || s[1][i][j] == ‘#‘)) s[0][i][j] = s[1][i][j] = ‘*‘;
else if(s[0][i][j] == ‘*‘ && s[1][i][j] == ‘#‘) s[1][i][j] = ‘*‘;
memset(vis, 0, sizeof vis);
vis[0][0][0] = 1;
bool ans = dfs(0, 0, 0, 0);
printf("%s\n", ans ? "YES" : "NO");
}
return 0;
}
标签：inline signed stdout 一个 bug gcd for comm tin
原文地址：http://www.cnblogs.com/dwtfukgv/p/7152646.html
