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\n",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\n",ans);
105 }
106 return 0;
107 }
end
时间： 2024-10-07 22:52:52