您所在位置：主题：[讨论]帮忙看一个算法题目，提交的时候出错
yg_sun
[专家分：160] 发布于 2008-05-16 23:25:00
第一次做ACM的题，也不太熟悉c++,提交后runtime error， 我试了几种组合，不管输出对不对，没有遇到什么exception，请帮忙看看吧。
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
// block stack will be stored using a list from bottom to top
//
// store block info
struct node
{
node* p;
node* n;
int num; // block num
};
// store command
struct command
{
int type;
int f;
int t;
command* next;
};
int* tbl; // if(the num block is on the bottom) tbl[num] = position; else tbl[num] = -1
node** idx; // idx[num] is a pointer to its node struct
command* head; // store command
void exec(int command, int f, int t)
{
if(f == t)
return;
// check the f and t is not on the same stack
node* cnode = idx[f]->n;
while(cnode != NULL)
{
if(cnode == idx[t])
return;
cnode = cnode->n;
}
cnode = idx[t]->n;
while(cnode != NULL)
{
if(cnode == idx[f])
return;
cnode = cnode->n;
}
switch(command)
{
case 0:
{
// move onto
node* ta = idx[t]->n; // blocks after t
node* fa = idx[f]->n; // blocks after f
idx[t]->n = idx[f]; // move f onto t
if(idx[f]->p != NULL)
idx[f]->p->n = idx[f]->n;
else // the block is on the bottom
{
if(fa != NULL) // the block after the current will become the bottom block
{
tbl[fa->num] = tbl[idx[f]->num];
fa->p = NULL;
}
tbl[idx[f]->num] = -1;
}
idx[f]->p = idx[t];
if(ta != NULL)
ta->p = idx[f];
idx[f]->n = ta;
break;
}
case 1:
// move over
{
node* fa = idx[f]->n;
node* tend = idx[t];
while(tend->n != NULL){tend = tend->n;};
tend->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p->n = idx[f]->n;
}
else
{
if(fa != NULL)
tbl[fa->num] = tbl[idx[f]->num];
tbl[idx[f]->num] = -1;
}
idx[f]->p = tend;
break;
}
case 2:
// pile onto
{
node* ta = idx[t]->n;
node* fend = idx[f];
idx[t]->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p = NULL;
}
else
{
tbl[idx[f]->num] = -1;
}
if(ta != NULL)
{
while(fend->n != NULL){fend = fend->n;};
fend->n = ta;
ta->p = fend;
}
break;
}
case 3:
// pile over
{
node* tend = idx[t];
while(tend->n != NULL){tend = tend->n;};
tend->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p->n = NULL;
}
else
{
tbl[idx[f]->num] = -1;
}
idx[f]->p = tend;
break;
}
}
}
void pr(int n)
{
node** tmp = new node*[n];
for(int m=0;m<n;m++)
tmp[m] = NULL;
for(int i=0;i<n;i++)
{
if(tbl[i] != -1)
tmp[tbl[i]] = idx[i];
}
for(int j=0;j<n;j++)
{
cout << j << ": ";
if(tmp[j]!=NULL)
{
node* _n = tmp[j];
while(_n)
{
cout << _n->num << " ";
_n = _n->n;
}
}
cout << endl;
}
delete tmp;
}
int main()
{
int n;
int k = 0;
string s;
cin >> n;
cin >> s;
head = new command;
head->type = -1;
command* cmd =head;
while(s != "quit")
{
cmd->next = new command;
cmd = cmd->next;
cmd->f = 0;
cmd->t = 0;
cmd->type = 0;
cmd->next = NULL;
if(s == "pile")
cmd->type = 2;
cin >> s;
istringstream iss1(s);
iss1 >> cmd->f;
cin >> s;
if(s == "over")
cmd->type += 1;
cin >> s;
istringstream iss2(s);
iss2 >> cmd->t;
cin >> s;
}
idx = new node*[n];
tbl = new int[n];
for(int i=0;i<10;i++)
{
tbl[i]=i;
idx[i] = new node;
idx[i]->num = i;
idx[i]->p = NULL;
idx[i]->n = NULL;
}
cmd = head->next;
while(cmd != NULL)
{
if(cmd->t < n && cmd->f < n)
exec(cmd->type, cmd->f, cmd->t);
cmd = cmd->next;
}
pr(n);
delete tbl;
for(int k1=0;k1<n;k1++)
delete idx[k1];
delete idx;
delete head;
return 0;
}
--------------------------------------------原题--------------------------------------
The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n-1) with block bi adjacent to block bi+1 for all 0<=i < n-1
The valid commands for the robot arm that manipulates blocks are:
* move a onto b
where a and b are block numbers, puts block a onto block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.
* move a over b
where a and b are block numbers, puts block a onto the top of the stack containing block b, after returning any blocks that are stacked on top of block a to their initial positions.
* pile a onto b
where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto block b. All blocks on top of block b are moved to their initial positions prior to the pile taking place. The blocks stacked above block a retain their order when moved.
* pile a over b
where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto the top of the stack containing block b. The blocks stacked above block a retain their original order when moved.
* quit
terminates manipulations in the block world.
Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.
The Input
The input begins with an integer n on a line by itself representing the number of blocks in the block world. You may assume that 0 < n < 25.
The number of blocks is followed by a sequence of block commands, one command per line. Your program should process all commands until the quit command is encountered.
You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.
The Output
The output should consist of the final state of the blocks world. Each original block position numbered i ( 0<=i<n where n is the number of blocks) should appear followed immediately by a colon. If there is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear stacked in that position with each block number separated from other block numbers by a space. Don't put any trailing spaces on a line.
There should be one line of output for each block position (i.e., n lines of output where n is the integer on the first line of input).
Sample Input
10
move 9 onto 1
move 8 over 1
move 7 over 1
move 6 over 1
pile 8 over 6
pile 8 over 5
move 2 over 1
move 4 over 9
quit
Sample Output
0: 0
1: 1 9 2 4
2:
3: 3
4:
5: 5 8 7 6
6:
7:
8:
9:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
// block stack will be stored using a list from bottom to top
//
// store block info
struct node
{
node* p;
node* n;
int num; // block num
};
// store command
struct command
{
int type;
int f;
int t;
command* next;
};
int* tbl; // if(the num block is on the bottom) tbl[num] = position; else tbl[num] = -1
node** idx; // idx[num] is a pointer to its node struct
command* head; // store command
void exec(int command, int f, int t)
{
if(f == t)
return;
// check the f and t is not on the same stack
node* cnode = idx[f]->n;
while(cnode != NULL)
{
if(cnode == idx[t])
return;
cnode = cnode->n;
}
cnode = idx[t]->n;
while(cnode != NULL)
{
if(cnode == idx[f])
return;
cnode = cnode->n;
}
switch(command)
{
case 0:
{
// move onto
node* ta = idx[t]->n; // blocks after t
node* fa = idx[f]->n; // blocks after f
idx[t]->n = idx[f]; // move f onto t
if(idx[f]->p != NULL)
idx[f]->p->n = idx[f]->n;
else // the block is on the bottom
{
if(fa != NULL) // the block after the current will become the bottom block
{
tbl[fa->num] = tbl[idx[f]->num];
fa->p = NULL;
}
tbl[idx[f]->num] = -1;
}
idx[f]->p = idx[t];
if(ta != NULL)
ta->p = idx[f];
idx[f]->n = ta;
break;
}
case 1:
// move over
{
node* fa = idx[f]->n;
node* tend = idx[t];
while(tend->n != NULL){tend = tend->n;};
tend->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p->n = idx[f]->n;
}
else
{
if(fa != NULL)
tbl[fa->num] = tbl[idx[f]->num];
tbl[idx[f]->num] = -1;
}
idx[f]->p = tend;
break;
}
case 2:
// pile onto
{
node* ta = idx[t]->n;
node* fend = idx[f];
idx[t]->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p = NULL;
}
else
{
tbl[idx[f]->num] = -1;
}
if(ta != NULL)
{
while(fend->n != NULL){fend = fend->n;};
fend->n = ta;
ta->p = fend;
}
break;
}
case 3:
// pile over
{
node* tend = idx[t];
while(tend->n != NULL){tend = tend->n;};
tend->n = idx[f];
if(idx[f]->p != NULL)
{
idx[f]->p->n = NULL;
}
else
{
tbl[idx[f]->num] = -1;
}
idx[f]->p = tend;
break;
}
}
}
void pr(int n)
{
node** tmp = new node*[n];
for(int m=0;m<n;m++)
tmp[m] = NULL;
for(int i=0;i<n;i++)
{
if(tbl[i] != -1)
tmp[tbl[i]] = idx[i];
}
for(int j=0;j<n;j++)
{
cout << j << ": ";
if(tmp[j]!=NULL)
{
node* _n = tmp[j];
while(_n)
{
cout << _n->num << " ";
_n = _n->n;
}
}
cout << endl;
}
delete tmp;
}
int main()
{
int n;
int k = 0;
string s;
cin >> n;
cin >> s;
head = new command;
head->type = -1;
command* cmd =head;
while(s != "quit")
{
cmd->next = new command;
cmd = cmd->next;
cmd->f = 0;
cmd->t = 0;
cmd->type = 0;
cmd->next = NULL;
if(s == "pile")
cmd->type = 2;
cin >> s;
istringstream iss1(s);
iss1 >> cmd->f;
cin >> s;
if(s == "over")
cmd->type += 1;
cin >> s;
istringstream iss2(s);
iss2 >> cmd->t;
cin >> s;
}
idx = new node*[n];
tbl = new int[n];
for(int i=0;i<10;i++)
{
tbl[i]=i;
idx[i] = new node;
idx[i]->num = i;
idx[i]->p = NULL;
idx[i]->n = NULL;
}
cmd = head->next;
while(cmd != NULL)
{
if(cmd->t < n && cmd->f < n)
exec(cmd->type, cmd->f, cmd->t);
cmd = cmd->next;
}
pr(n);
delete tbl;
for(int k1=0;k1<n;k1++)
delete idx[k1];
delete idx;
delete head;
return 0;
}
--------------------------------------------原题--------------------------------------
The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n-1) with block bi adjacent to block bi+1 for all 0<=i < n-1
The valid commands for the robot arm that manipulates blocks are:
* move a onto b
where a and b are block numbers, puts block a onto block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.
* move a over b
where a and b are block numbers, puts block a onto the top of the stack containing block b, after returning any blocks that are stacked on top of block a to their initial positions.
* pile a onto b
where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto block b. All blocks on top of block b are moved to their initial positions prior to the pile taking place. The blocks stacked above block a retain their order when moved.
* pile a over b
where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto the top of the stack containing block b. The blocks stacked above block a retain their original order when moved.
* quit
terminates manipulations in the block world.
Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.
The Input
The input begins with an integer n on a line by itself representing the number of blocks in the block world. You may assume that 0 < n < 25.
The number of blocks is followed by a sequence of block commands, one command per line. Your program should process all commands until the quit command is encountered.
You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.
The Output
The output should consist of the final state of the blocks world. Each original block position numbered i ( 0<=i<n where n is the number of blocks) should appear followed immediately by a colon. If there is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear stacked in that position with each block number separated from other block numbers by a space. Don't put any trailing spaces on a line.
There should be one line of output for each block position (i.e., n lines of output where n is the integer on the first line of input).
Sample Input
10
move 9 onto 1
move 8 over 1
move 7 over 1
move 6 over 1
pile 8 over 6
pile 8 over 5
move 2 over 1
move 4 over 9
quit
Sample Output
0: 0
1: 1 9 2 4
2:
3: 3
4:
5: 5 8 7 6
6:
7:
8:
9: