Wang Xifeng's Little Plot http://acm.hdu.edu.cn/showproblem.php?pid=5024
预处理出每个点八个方向能走的最远距离，然后枚举起点，枚举方向，每走一步都要枚举左转和右转的情况，因为预处理好了，所以可以直接算出来。
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 const int M=128; 5 char a[M][M]; 6 int n,dp[M][M][8]; 7 int dx[]={-1,-1,0,1,1,1,0,-1}; 8 int dy[]={0,1,1,1,0,-1,-1,-1}; 9 bool inside(const int &x,const int &y){ 10 if(x>=0&&x<n&&y>=0&&y<n) return true;return false; 11 } 12 int getfar(int x,int y,int dir){ 13 int res=0; 14 while(true){ 15 x+=dx[dir]; 16 y+=dy[dir]; 17 if(!inside(x,y)||a[x][y]=='#') break; 18 res++; 19 } 20 return res; 21 } 22 int turnleft(int dir){ 23 return (dir+6)%8; 24 } 25 int turnright(int dir){ 26 return (dir+2)%8; 27 } 28 int solve(int x,int y,int dir){ 29 int res=1,step=1,left=turnleft(dir),right=turnright(dir); 30 while(true){ 31 x+=dx[dir]; 32 y+=dy[dir]; 33 if(!inside(x,y)||a[x][y]=='#') break; 34 step++; 35 res=max(res,step+dp[x][y][left]); 36 res=max(res,step+dp[x][y][right]); 37 } 38 return res; 39 } 40 int main(){ 41 while(~scanf("%d",&n),n){ 42 for(int i=0;i<n;i++){ 43 scanf("%s",a[i]); 44 } 45 for(int i=0;i<n;i++){ 46 for(int j=0;j<n;j++){ 47 for(int k=0;k<8;k++){ 48 if(a[i][j]=='#'){ 49 dp[i][j][k]=-1; 50 } 51 else{ 52 dp[i][j][k]=getfar(i,j,k); 53 } 54 } 55 } 56 } 57 int ans=0; 58 for(int i=0;i<n;i++){ 59 for(int j=0;j<n;j++){ 60 if(a[i][j]=='.'){ 61 for(int k=0;k<8;k++){ 62 int now=solve(i,j,k); 63 ans=max(ans,now); 64 } 65 } 66 } 67 } 68 printf("%d ",ans); 69 } 70 return 0; 71 }
A Corrupt Mayor's Performance Art http://acm.hdu.edu.cn/showproblem.php?pid=5023
成段更新。
1 #include<cstdio> 2 #include<cstring> 3 #define mt(a,b) memset(a,b,sizeof(a)) 4 #define lrrt int L,int R,int rt 5 #define iall 1,n,1 6 #define imid int mid=(L+R)>>1 7 #define lson L,mid,rt<<1 8 #define rson mid+1,R,rt<<1|1 9 const int M=1000010; 10 struct T{ 11 int cover,lazy; 12 }tree[M<<2]; 13 void build(lrrt){ 14 tree[rt].cover=2; 15 tree[rt].lazy=0; 16 if(L==R) return ; 17 imid; 18 build(lson); 19 build(rson); 20 } 21 void pushdown(int rt){ 22 if(tree[rt].lazy){ 23 tree[rt<<1].lazy=tree[rt<<1|1].lazy=tree[rt<<1].cover=tree[rt<<1|1].cover=tree[rt].lazy; 24 tree[rt].lazy=0; 25 } 26 } 27 void pushup(int rt){ 28 tree[rt].cover=tree[rt<<1].cover==tree[rt<<1|1].cover?tree[rt<<1].cover:0; 29 } 30 void update(int x,int y,int z,lrrt){ 31 if(x<=L&&R<=y){ 32 tree[rt].cover=tree[rt].lazy=z; 33 return; 34 } 35 imid; 36 pushdown(rt); 37 if(mid>=x) update(x,y,z,lson); 38 if(mid<y) update(x,y,z,rson); 39 pushup(rt); 40 } 41 bool vis[32]; 42 void query(int x,int y,lrrt){ 43 if(L==R){ 44 vis[tree[rt].cover]=true; 45 return ; 46 } 47 imid; 48 if(x<=L&&R<=y){ 49 if(tree[rt].cover){ 50 vis[tree[rt].cover]=true; 51 return ; 52 } 53 pushdown(rt); 54 query(x,y,lson); 55 query(x,y,rson); 56 return ; 57 } 58 pushdown(rt); 59 if(mid>=x) query(x,y,lson); 60 if(mid<y) query(x,y,rson); 61 } 62 int main(){ 63 int n,m,x,y,z; 64 char op[4]; 65 while(~scanf("%d%d",&n,&m),n|m){ 66 build(iall); 67 while(m--){ 68 scanf("%s%d%d",op,&x,&y); 69 if(op[0]=='P'){ 70 scanf("%d",&z); 71 update(x,y,z,iall); 72 } 73 else{ 74 mt(vis,0); 75 query(x,y,iall); 76 bool flag=false; 77 for(int i=1;i<=30;i++){ 78 if(vis[i]){ 79 if(flag) printf(" "); 80 printf("%d",i); 81 flag=true; 82 } 83 } 84 puts(""); 85 } 86 } 87 } 88 return 0; 89 }
Saving Tang Monk http://acm.hdu.edu.cn/showproblem.php?pid=5025
集齐m把钥匙，就可以拯救唐僧，没集齐也可以从他身上踩过去，拿钥匙必须从小到大拿，蛇只要杀一次，所以要记录钥匙的状态和蛇是否被杀的状态。bfs。
1 #include<cstdio> 2 #include<cstring> 3 #include<cctype> 4 #include<queue> 5 #define mt(a,b) memset(a,b,sizeof(a)) 6 using namespace std; 7 const int M=128; 8 char a[M][M]; 9 int n,m; 10 struct Snake{ 11 int x,y; 12 }ts[8]; 13 struct Q{ 14 int step,sta,x,y,snake; 15 friend bool operator <(const Q &a,const Q &b){ 16 return a.step>b.step; 17 } 18 }now,pre; 19 priority_queue<Q> q; 20 bool vis[M][M][1<<9]; 21 int dx[]={0,0,1,-1}; 22 int dy[]={1,-1,0,0}; 23 int getkey(const int &psta,const int &x,const int &y){ 24 int ressta=psta; 25 if(isdigit(a[x][y])){ 26 int his=a[x][y]-'0'; 27 his=1<<(his-1); 28 if(his-1==psta){ 29 ressta|=his; 30 } 31 } 32 return ressta; 33 } 34 int bfs(){ 35 int ls=0; 36 for(int i=0;i<n;i++){ 37 for(int j=0;j<n;j++){ 38 if(a[i][j]=='K'){ 39 now.x=i; 40 now.y=j; 41 } 42 if(a[i][j]=='S'){ 43 ts[ls].x=i; 44 ts[ls].y=j; 45 ls++; 46 } 47 } 48 } 49 now.snake=0; 50 now.sta=0; 51 now.step=0; 52 mt(vis,0); 53 vis[now.x][now.y][now.sta]=true; 54 int End=(1<<m)-1; 55 while(!q.empty()) q.pop(); 56 q.push(now); 57 while(!q.empty()){ 58 pre=q.top(); 59 q.pop(); 60 if(a[pre.x][pre.y]=='T'&&pre.sta==End) return pre.step; 61 for(int i=0;i<4;i++){ 62 int tx=pre.x+dx[i]; 63 int ty=pre.y+dy[i]; 64 if(tx>=0&&tx<n&&ty>=0&&ty<n&&a[tx][ty]!='#'){ 65 now.sta=getkey(pre.sta,tx,ty); 66 if(!vis[tx][ty][now.sta]){ 67 vis[tx][ty][now.sta]=true; 68 now.x=tx; 69 now.y=ty; 70 now.snake=pre.snake; 71 if(a[tx][ty]=='S'){ 72 int id=0; 73 for(int j=0;j<ls;j++){ 74 if(ts[j].x==tx&&ts[j].y==ty){ 75 id=j; 76 break; 77 } 78 } 79 if((now.snake>>id)&1){ 80 now.step=pre.step+1; 81 } 82 else{ 83 now.snake|=(1<<id); 84 now.step=pre.step+2; 85 } 86 } 87 else{ 88 now.step=pre.step+1; 89 } 90 q.push(now); 91 } 92 } 93 } 94 } 95 return -1; 96 } 97 int main(){ 98 while(~scanf("%d%d",&n,&m),n|m){ 99 for(int i=0;i<n;i++){ 100 scanf("%s",a[i]); 101 } 102 int ans=bfs(); 103 if(ans==-1) puts("impossible"); 104 else printf("%d ",ans); 105 } 106 return 0; 107 }
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