[BFS/DFS]HOJ2581Go

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Go

My Tags (Edit)
 Source : Stanford Programming Contest 2007 Time limit : 1 sec Memory limit : 64 M

Submitted : 201, Accepted : 117

In the game of Go, two players alternate placing black and white stones on lattice points of an n * n grid, each attempting to surround as much territory(i.e., regions of unfilled lattice points) as possible. At the end of the game, the score for each player is the total area of the territory surrounded by his or her stones. Given the locations of black and white stones on a Go board at the end of a match, your task is to compute the score of each player in order to determine the winner.

Formally, two grid lattice points with coordinates (r, c) and (r', c') are adjacent if |r - r'| + |c - c'| = 1. A connected region of unfilled lattice points belongs to one player's territory if all adjacent filled lattice points contain stones belonging to that player(see Figure 1). Finally, a player's score consists of the number of unfilled lattice points in his or her territory.

Figure 1: Diagram of a 9 * 9 Go board. Unfilled lattice points belonging to black's territory are marked with B, and unfilled lattice points belonging to white's territory are marked with W. Neutral unfilled lattice points are unmarked. In the game above, white wins by 21 - 3 = 18.

Input

The input test file will contain multiple cases, each consisting of three lines. Each test case begins with a line containing three integers, n(1 ≤ n ≤ 19), b and w(b ≥ 0, w ≥ 0 and 1 ≤ b + w ≤ n2). Here, n denotes the size of the board, b is the number of black pieces placed, and w is the number of white pieces placed. The second line of each test case contains b pairs of integers r1 c1 . . . rb cb (where 1 ≤ ri, ci ≤ n) indicating the positions of the b black stones. The third line of each test case contains w pairs of integers r'1 c'1 ... r'w c'w(1 ≤ r'i, c'i ≤ n) indicating the positions of the w white stones. No two stones will be located at the same lattice point. Input is terminated by a single line containing only the number 0; do not process this line.

Output

For each test case, print either "White wins by ____", "Black wins by ____", or "Draw".

Sample Input

```1 1 0
1 1
2 0 1
1 1
5 12 4
1 1 1 2 1 3 2 1 2 3 3 1 3 3 4 1 4 3 5 1 5 2 5 3
1 4 2 4 3 4 3 5
0```

Sample Output

```Draw
White wins by 3
Black wins by 1```

DFS：

```#include<iostream>
#include<stack>
#include<cstdio>
#include<cstring>
using namespace std;
int maps[25][25];
struct node{
int x,y;
}st;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int flagb,flagw,numb,numw,num;
void dfs(node st,int n){
stack<node> q;
int i;
node a,b;
while(!q.empty()){
q.pop();
}
flagb=0;flagw=0;num=1;
q.push(st);
maps[st.x][st.y]=3;
while(!q.empty()){
a=q.top();
q.pop();
for(i=0;i<4;i++){
b.x=a.x+dx[i];
b.y=a.y+dy[i];
if(b.x>0&&b.x<=n&&b.y>0&&b.y<=n){
if(maps[b.x][b.y]==0){
maps[b.x][b.y]=3;
q.push(b);
num++;
}
else if(maps[b.x][b.y]==1)
flagb=1;
else if(maps[b.x][b.y]==2)
flagw=1;
}
}
}
}
int main(){
int n,b,w,i,j,x,y;
while(scanf("%d%d%d",&n,&b,&w)==3){
flagb=0;flagw=0;numb=0;numw=0;num=0;
memset(maps,0,sizeof(maps));
//黑子为1
for(i=1;i<=b;i++){
scanf("%d%d",&x,&y);
maps[x][y]=1;
}
//白子为2
for(i=1;i<=w;i++){
scanf("%d%d",&x,&y);
maps[x][y]=2;
}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(maps[i][j]==0){
st.x=i;
st.y=j;
dfs(st,n);
if(flagb==0&&flagw==1)
numw+=num;
if(flagb==1&&flagw==0)
numb+=num;
}
}
}
if(numw==numb)
printf("Draw\n");
else if(numw>numb)
printf("White wins by %d\n",numw-numb);
else
printf("Black wins by %d\n",numb-numw);
}
return 0;
}
```

BFS:

```#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
int maps[25][25];
struct node{
int x,y;
}st;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int flagb,flagw,numb,numw,num;
void bfs(node st,int n){
queue<node> q;
int i;
node a,b;
while(!q.empty()){
q.pop();
}
flagb=0;flagw=0;num=1;
q.push(st);
maps[st.x][st.y]=3;
while(!q.empty()){
a=q.front();
q.pop();
for(i=0;i<4;i++){
b.x=a.x+dx[i];
b.y=a.y+dy[i];
if(b.x>0&&b.x<=n&&b.y>0&&b.y<=n){
if(maps[b.x][b.y]==0){
maps[b.x][b.y]=3;
q.push(b);
num++;
}
else if(maps[b.x][b.y]==1)
flagb=1;
else if(maps[b.x][b.y]==2)
flagw=1;
}
}
}
}
int main(){
int n,b,w,i,j,x,y;
while(scanf("%d%d%d",&n,&b,&w)==3){
flagb=0;flagw=0;numb=0;numw=0;num=0;
memset(maps,0,sizeof(maps));
//黑子为1
for(i=1;i<=b;i++){
scanf("%d%d",&x,&y);
maps[x][y]=1;
}
//白子为2
for(i=1;i<=w;i++){
scanf("%d%d",&x,&y);
maps[x][y]=2;
}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(maps[i][j]==0){
st.x=i;
st.y=j;
bfs(st,n);
if(flagb==0&&flagw==1)
numw+=num;
if(flagb==1&&flagw==0)
numb+=num;
}
}
}
if(numw==numb)
printf("Draw\n");
else if(numw>numb)
printf("White wins by %d\n",numw-numb);
else
printf("Black wins by %d\n",numb-numw);
}
return 0;
}```

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Go

My Tags (Edit)
 Source : Stanford Programming Contest 2007 Time limit : 1 sec Memory limit : 64 M

Submitted : 201, Accepted : 117

In the game of Go, two players alternate placing black and white stones on lattice points of an n * n grid, each attempting to surround as much territory(i.e., regions of unfilled lattice points) as possible. At the end of the game, the score for each player is the total area of the territory surrounded by his or her stones. Given the locations of black and white stones on a Go board at the end of a match, your task is to compute the score of each player in order to determine the winner.

Formally, two grid lattice points with coordinates (r, c) and (r', c') are adjacent if |r - r'| + |c - c'| = 1. A connected region of unfilled lattice points belongs to one player's territory if all adjacent filled lattice points contain stones belonging to that player(see Figure 1). Finally, a player's score consists of the number of unfilled lattice points in his or her territory.

Figure 1: Diagram of a 9 * 9 Go board. Unfilled lattice points belonging to black's territory are marked with B, and unfilled lattice points belonging to white's territory are marked with W. Neutral unfilled lattice points are unmarked. In the game above, white wins by 21 - 3 = 18.

Input

The input test file will contain multiple cases, each consisting of three lines. Each test case begins with a line containing three integers, n(1 ≤ n ≤ 19), b and w(b ≥ 0, w ≥ 0 and 1 ≤ b + w ≤ n2). Here, n denotes the size of the board, b is the number of black pieces placed, and w is the number of white pieces placed. The second line of each test case contains b pairs of integers r1 c1 . . . rb cb (where 1 ≤ ri, ci ≤ n) indicating the positions of the b black stones. The third line of each test case contains w pairs of integers r'1 c'1 ... r'w c'w(1 ≤ r'i, c'i ≤ n) indicating the positions of the w white stones. No two stones will be located at the same lattice point. Input is terminated by a single line containing only the number 0; do not process this line.

Output

For each test case, print either "White wins by ____", "Black wins by ____", or "Draw".

Sample Input

```1 1 0
1 1
2 0 1
1 1
5 12 4
1 1 1 2 1 3 2 1 2 3 3 1 3 3 4 1 4 3 5 1 5 2 5 3
1 4 2 4 3 4 3 5
0```

Sample Output

```Draw
White wins by 3
Black wins by 1```

DFS：

```#include<iostream>
#include<stack>
#include<cstdio>
#include<cstring>
using namespace std;
int maps[25][25];
struct node{
int x,y;
}st;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int flagb,flagw,numb,numw,num;
void dfs(node st,int n){
stack<node> q;
int i;
node a,b;
while(!q.empty()){
q.pop();
}
flagb=0;flagw=0;num=1;
q.push(st);
maps[st.x][st.y]=3;
while(!q.empty()){
a=q.top();
q.pop();
for(i=0;i<4;i++){
b.x=a.x+dx[i];
b.y=a.y+dy[i];
if(b.x>0&&b.x<=n&&b.y>0&&b.y<=n){
if(maps[b.x][b.y]==0){
maps[b.x][b.y]=3;
q.push(b);
num++;
}
else if(maps[b.x][b.y]==1)
flagb=1;
else if(maps[b.x][b.y]==2)
flagw=1;
}
}
}
}
int main(){
int n,b,w,i,j,x,y;
while(scanf("%d%d%d",&n,&b,&w)==3){
flagb=0;flagw=0;numb=0;numw=0;num=0;
memset(maps,0,sizeof(maps));
//黑子为1
for(i=1;i<=b;i++){
scanf("%d%d",&x,&y);
maps[x][y]=1;
}
//白子为2
for(i=1;i<=w;i++){
scanf("%d%d",&x,&y);
maps[x][y]=2;
}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(maps[i][j]==0){
st.x=i;
st.y=j;
dfs(st,n);
if(flagb==0&&flagw==1)
numw+=num;
if(flagb==1&&flagw==0)
numb+=num;
}
}
}
if(numw==numb)
printf("Draw\n");
else if(numw>numb)
printf("White wins by %d\n",numw-numb);
else
printf("Black wins by %d\n",numb-numw);
}
return 0;
}
```

BFS:

```#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
int maps[25][25];
struct node{
int x,y;
}st;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int flagb,flagw,numb,numw,num;
void bfs(node st,int n){
queue<node> q;
int i;
node a,b;
while(!q.empty()){
q.pop();
}
flagb=0;flagw=0;num=1;
q.push(st);
maps[st.x][st.y]=3;
while(!q.empty()){
a=q.front();
q.pop();
for(i=0;i<4;i++){
b.x=a.x+dx[i];
b.y=a.y+dy[i];
if(b.x>0&&b.x<=n&&b.y>0&&b.y<=n){
if(maps[b.x][b.y]==0){
maps[b.x][b.y]=3;
q.push(b);
num++;
}
else if(maps[b.x][b.y]==1)
flagb=1;
else if(maps[b.x][b.y]==2)
flagw=1;
}
}
}
}
int main(){
int n,b,w,i,j,x,y;
while(scanf("%d%d%d",&n,&b,&w)==3){
flagb=0;flagw=0;numb=0;numw=0;num=0;
memset(maps,0,sizeof(maps));
//黑子为1
for(i=1;i<=b;i++){
scanf("%d%d",&x,&y);
maps[x][y]=1;
}
//白子为2
for(i=1;i<=w;i++){
scanf("%d%d",&x,&y);
maps[x][y]=2;
}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(maps[i][j]==0){
st.x=i;
st.y=j;
bfs(st,n);
if(flagb==0&&flagw==1)
numw+=num;
if(flagb==1&&flagw==0)
numb+=num;
}
}
}
if(numw==numb)
printf("Draw\n");
else if(numw>numb)
printf("White wins by %d\n",numw-numb);
else
printf("Black wins by %d\n",numb-numw);
}
return 0;
}```