标签：style blog io sp for on 2014 log ad
状态压缩的题，第一次WA了，怎么改都不对，交了十几遍，之后重新写了一个，1A了
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 51;
const int dir[4][2] = {{0,1},{-1,0},{0,-1},{1,0}};
int n,m,t;
int door[maxn][maxn][4];
int key[maxn][maxn];
int vis[maxn][maxn][(1 << 11)];
struct St{
int x,y,key,step;
}now;
void init(){
memset(key,0,sizeof(key));
memset(door,-1,sizeof(door));//-1代表可以任意的通过
memset(vis,0,sizeof(vis));
}
void add_door(int x1,int y1,int x2,int y2,int op){
if(x1 == x2){
if(y1 > y2){
door[x1][y1][2] = (1 << op);
door[x2][y2][0] = (1 << op);
}
else{
door[x1][y1][0] = (1 << op);
door[x2][y2][2] = (1 << op);
}
}
else{
if(x1 > x2){
door[x1][y1][1] = (1 << op);
door[x2][y2][3] = (1 << op);
}
else{
door[x1][y1][3] = (1 << op);
door[x2][y2][1] = (1 << op);
}
}
}
bool judge(int d){
if(door[now.x][now.y][d] == -1) return true; //可以直接通过
if(door[now.x][now.y][d] == 1) return false; //是一面墙
int op = door[now.x][now.y][d]&now.key;
if(!op) return false;
return true;
}
void dfs(){
now.x = 1; now.y = 1; now.key = key[1][1]; now.step = 0;
vis[now.x][now.y][now.key] = 1;
queue<St>q;
while(!q.empty()) q.pop();
q.push(now);
while(!q.empty()){
now = q.front(); q.pop();
//printf("%d %d\n",now.x,now.y);;
if(now.x == n && now.y == m){
printf("%d\n",now.step);
return;
}
for(int d = 0; d < 4; d++){
int x = now.x + dir[d][0];
int y = now.y + dir[d][1];
if(x >= 1 && x <= n && y >= 1 && y <= m){
if(!judge(d)) continue;
int c_key = now.key|key[x][y];
if(!vis[x][y][c_key]){
St temp; temp.x = x; temp.y = y; temp.key = c_key;
temp.step = now.step + 1;
q.push(temp);
vis[x][y][c_key] = 1;
}
}
}
}
printf("-1\n");
}
int main(){
while(scanf("%d%d%d",&n,&m,&t) != EOF){
int a,b;
init();
scanf("%d",&a);
for(int i = 0; i < a; i++){
int x1,y1,x2,y2,op;
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&op);
add_door(x1,y1,x2,y2,op);
}
scanf("%d",&b);
for(int i = 0; i < b; i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
key[x][y] |= (1 << z); //一个坐标的钥匙,状态压缩
}
dfs();
}
return 0;
}
标签：style blog io sp for on 2014 log ad
原文地址：http://blog.csdn.net/u013451221/article/details/40931661
