#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int prime[40]={0,0,1},vis[40],num[40],n;
void checkPrime(){
for(int i=3;i<=40;i++){
prime[i]=1;
for(int k=2;k<i;k++){
if(i%k==0){
prime[i]=0;
break;
}
}
}
}
void DFS(int i){
if(i==n+1) {
printf("1");
for(int i=2;i<=n;i++)
printf(" %d",num[i]);
printf("\n");
return;
}
for(int c=2;c<=n;c++){
if(vis[c]==0 && prime[num[i-1]+c]){
if(i==n&&prime[c+1]!=1) return;
vis[c]=1;
num[i]=c;
DFS(i+1);
vis[c]=0;
}
}
}
int main(){
int t=0;
checkPrime();
while(~scanf("%d",&n)){
printf("Case %d:\n",++t);
memset(vis, 0, sizeof(vis));
memset(num, 0, sizeof(num));
vis[1]=1;
num[1]=1;
num[n+1]=1;
DFS(2);
printf("\n");
}
return 0;
}
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
using namespace std;
int m,n,sum,vis[105][105];
char map[105][105];
int dx[3]={0,1,-1};
int dy[3]={0,1,-1};
void DFS(int x,int y,int f){
if(map[x][y]=='*' || vis[x][y]) return;
if(f==0) sum++;
vis[x][y]=1;
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
int fx = x+dx[i];
int fy = y+dy[j];
if(fx>=0&&fx<m&&fy>=0&&fy<n&&map[fx][fy]=='@'){
DFS(fx,fy,1);
}
}
}
}
int main(){
//freopen("cin.in", "r", stdin);
while (~scanf("%d %d",&m,&n)&&m) {
for(int i=0;i<m;i++) scanf("%s",map[i]);
memset(vis, 0, sizeof(vis));
sum=0;
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
if(map[i][j]=='@')
DFS(i,j,0);
printf("%d\n",sum);
}
return 0;
}
这题有个难点是怎么判断是不是同一个油田,我选择在dfs里传入参数f,如果是通过遍历进入的f为0,通过dfs进入的为1,这样就很好的区分了油田。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int N,START,END,step[205],vis[205];
struct Node{
int num,step;
};
int BFS(){
Node node;
node.num = START;
node.step = 0;
vis[START]=1;
queue<Node> q;
q.push(node);
while (!q.empty()) {
Node now =q.front();
q.pop();
if(now.num == END) return now.step;
for(int i=-1;i<=1;i+=2){
Node s;
s.num = now.num+i*step[now.num];
if(s.num>0&&vis[s.num]==0){
s.step = now.step+1;
vis[s.num]=1;
q.push(s);
}
}
}
return -1;
}
int main(){
while (~scanf("%d",&N)&&N) {
scanf("%d%d",&START,&END);
for(int i=1;i<=N;i++) scanf("%d",&step[i]);
memset(vis, 0, sizeof(vis));
printf("%d\n",BFS());
}
return 0;
}