您的位置:首页 > 其它

HDU-OJ 杭电1010

2018-01-31 00:54 387 查看
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;

char a[9][9];
int table[51][51];
int visited[50];
int T;
int final = 0;
int n, m;
int minPath;
int xStart, yStart;
int xEnd, yEnd;

void dfs(int s,int degree){
degree++;

if(visited[s] == 1)
return ;
if(final == 1) return ;

visited[s] = 1;
if(degree == T && a[(s-1)/m+1][(s-1)%m+1]=='D') final = 1;

if(T - degree < abs(xEnd - ((s-1)/m+1)) + abs(yEnd - ((s-1)%m+1))){
//        cout<<abs(xEnd - (s/(m+1)+1))<<" "<<abs(yEnd - (s - s/(m+1)*4));
visited[s] = 0;
return ;
}
//    cout<<"X="<<s/m + 1<<","<<"Y="<<s%m<<","<<"degree="<<degree<<endl;
//查4次
{
int i=s-1;
int v=table[s][i];
if(v==1)
dfs(i,degree);

}
{
int i=s+1;
int v=table[s][i];
if(v==1)
dfs(i,degree);

}
{
int i=s-m;
int v=table[s][i];
if(v==1)
dfs(i,degree);

}
{
int i=s+m;
int v=table[s][i];
if(v==1)
dfs(i,degree);

}

visited[s] = 0;
}

int main(){
int Start;
int end;
int count;

//initialize

while(scanf("%d%d%d",&n,&m,&T) != EOF && n != 0 && m != 0 && T != 0){
//初始化
for(int i = 0; i < 9; i++)
for(int j = 0; j < 9; j++)
a[i][j] = 'X';
memset(table,0,sizeof(table));
memset(visited,0,sizeof(visited));
final = 0;
count = 0;

for(int i = 1; i <= n; i++){
getchar();
for(int j = 1; j <= m; j++){
scanf("%c",&a[i][j]);
}
}
getchar();
//test
//        for(int i = 1; i <= n; i++){
//            for(int j = 1; j <= m; j++){
//                cout<<a[i][j]<<" ";
//            }
//            cout <<endl;
//        }
//
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
if(a[i][j] == 'S'){
Start = (i-1)*m + j;
xStart = i;
yStart = j;
}
if(a[i][j] == 'D'){
end = (i-1)*m + j;
xEnd = i;
yEnd = j;
}
}
}

//Create a adjacency list
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
count++;
if(a[i][j] == 'X') continue;
if(a[i][j-1]=='.'||a[i][j-1]=='D') {
table[count][count-1] = 1;
}
if(a[i][j+1]=='.'||a[i][j+1]=='D') {
table[count][count+1] = 1;
}
if(a[i-1][j]=='.'||a[i-1][j]=='D') {
table[count][count-m] = 1;
}
if(a[i+1][j]=='.'||a[i+1][j]=='D') {
table[count][count+m] = 1;
}

}
}

//        for(int i = 1; i <= n*m; i++){
//            for(int j = 1; j <= m*n; j++){
//                cout<<table[i][j]<<" ";
//            }
//            cout<<endl;
//        }
if(T < abs(xStart - xEnd) + abs(yStart - yEnd))    cout<<"NO\n";
else if((T - abs(xStart - xEnd) - abs(yStart - yEnd)) & 1){
cout<<"NO\n";
}
else{
dfs(Start,-1);
if(final == 1) cout <<"YES"<<endl;
else cout <<"NO"<<endl;
}
}
return 0;
}


这题风格太差 没必要邻接矩阵!很容易超时!!

下面介绍一下,我校ACM大神的算法

#include "iostream"
#include "cstring"
using namespace std;
const int maxn = 13;
char a[maxn][maxn];
int vis[maxn][maxn];

int T, n, m, final, sx, sy, ex, ey;
int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
void dfs(int x, int y, int step){
if (step > T) return;
if (a[x][y] == 'D' && step == T) final = 1;
if (final) return;
int left = T - step, les = abs(x - ex) + abs(y - ey);
if (left < les) return;
if (left % 2 != les % 2) return;

for (int i = 0; i < 4; i++) {
int nx = x + dx[i], ny = y + dy[i];
if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
if (a[nx][ny] == 'X') continue;
if (vis[nx][ny]) continue;
vis[nx][ny] = 1;
dfs(nx, ny, step + 1);
vis[nx][ny] = 0;
}
}

int main(){
#ifdef AC
freopen("data.txt", "r", stdin);
#endif
while(scanf("%d%d%d", &n, &m, &T) != EOF && n){
final = 0;
for(int i = 0; i < n; i++) scanf("%s", a[i]);
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {

94ab
if (a[i][j] == 'S') sx = i, sy = j, vis[i][j] = 1;
else if (a[i][j] == 'D') ex = i, ey = j, vis[i][j] = 0;
else if (a[i][j] == 'X') vis[i][j] = 1;
else vis[i][j] = 0;
}
dfs(sx, sy, 0);
if(final) puts("YES");
else puts("NO");

}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: