Codeforces 724C Ray Tracing 模拟
题意：给你一张n*m的图，横坐标从0到m，纵坐标从0到n，直线y=n，y=0，x=0，x=m都是墙，现有光束从0,0出发向1，1射去，每进过一个点耗费1秒，当遇到墙角时，光束停止，当射到墙面上时，反射，入射角等于出射角，先有k个传感器在不同的点上，问光束第一次经过这些点的时间，若不经过输出-1。
题解：这题真是卡了我半天，真的是半天，我先按点一个一个搜t在test7，再按边搜然后对于每条边找一遍有没有传感器在上面，又t在了8，当然啦，这样的姿势一看就是不正确的，然而蒟蒻的我还是连踩两发坑，那么正确的思路应该是先求出每个传感器所在直线y=x+b,y=-x+b的截距。然后按边搜，找出每条边的截距，然后再通过截距找对应的传感器.
下面是正确的代码。
#include
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
const LL INF=0x3f3f3f3f3f3f3f3fll;
struct node {
int x,y;
LL step;
int dir;
int id;
};
int vis[MAXN];
typedef pair PAIR;
LL ans[MAXN];
map > >sz,sf;
void judge(node now) {
int b;
int flag=0;
if (now.dir==1||now.dir==2) b=now.y-now.x;
else b=now.x+now.y,flag=1;
if (flag==1) {
for (auto it :sf[b]) {
int rx=it.first,id=it.second;
ans[id]=min(ans[id],now.step+abs(now.x-rx));
}
sf.erase(b);
}
else {
for (auto it : sz[b]) {
int rx=it.first,id=it.second;
ans[id]=min(ans[id],now.step+abs(now.x-rx));
}
sz.erase(b);
}
}
int main() {
int n,m,k;
scanf("%d %d %d",&m,&n,&k);
int x,y;
for (int i=1; i<=k; ++i) {
scanf("%d %d",&x,&y);
swap(x,y);
sz[y-x].push_back(PAIR(x,i));
sf[y+x].push_back(PAIR(x,i));
}
MEMINF(ans);
node now;
now.x=0,now.y=0,now.dir=1,now.step=0;
int mx,my;
bool flag=false;
int cnt=0;
while (true) {
node lim;
if (now.dir==1) {
int r=m-now.y,u=n-now.x;
if (u 接下来是两发t的，写了比较长时间不舍得删掉，哈哈。
#include
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
struct node {
int x,y;
int step;
int dir;
int id;
};
int dx[4]={1,1,-1,-1};
template
inline bool read(T &n)
{
T x = 0, tmp = 1;
char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
int dy[4]={-1,1,-1,1};
int vis[MAXN];
typedef pair PAIR;
typedef map mpp;
mpp mp;
queueq;
int qsize;
double k1,k2;
void judge(node u,node lim) {
int k=qsize;
for(int i=0; i1e5) break;
if (now.dir==1) {
int r=m-now.y,u=n-now.x;
if (u
#include
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
struct node {
int x,y;
int step;
int dir;
};
int dx[4]={1,1,-1,-1};
int dy[4]={-1,1,-1,1};
int ans[MAXN];
bool vis[MAXN];
typedef pair PAIR;
typedef map mpp;
mpp mp;
int main() {
int n,m,k;
scanf("%d %d %d",&n,&m,&k);
int x,y;
for (int i=1; i<=k; ++i) {
scanf("%d %d",&x,&y);
mp[PAIR(x,y)]=i;
}
node pre,now;
MEM(vis,0);
MEM(ans,0);
now.x=1,now.y=1,now.step=1,now.dir=1;
int flag=0;
int cnt=0;
while(true) {
mpp::iterator it;
it=mp.find(PAIR(now.x,now.y));
int v=it->second;
if (v&&!vis[v]) {
ans[v]=now.step;
vis[v]=1;
flag++;
}
if (cnt>1e5)break;
if (flag==k) break;
now.x+=dx[now.dir];
now.y+=dy[now.dir];
now.step++;
if (now.x==0&&now.y==0)break;
if (now.x==n&&now.y==m) break;
if (now.x==0&&now.y==m) break;
if (now.x==n&&now.y==m)break;
if (now.x==0) {
if (now.dir==3) now.dir=1;
else if (now.dir==2) now.dir=0;
}
else if (now.x==n) {
if (now.dir==1) now.dir=3;
else if (now.dir==0) now.dir=2;
}
else if (now.y==0) {
if (now.dir==2) now.dir=3;
else if (now.dir==0) now.dir=1;
}
else if (now.y==m) {
if (now.dir==3) now.dir=2;
else if (now.dir==1) now.dir=0;
}
}
for (int i=1; i<=k; ++i) {
if (ans[i]==0)puts("-1");
else printf("%d\n",ans[i]);
}
}