poj 1564
给出一个s,n个数，输出所有的能够得到s的方案
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 int a[15]; 8 int ans[15]; 9 int n,s,flag; 10 11 void dfs(int x,int sum,int cnt){ 12 if(sum > s) return; 13 if(sum == s){ 14 flag = 1; 15 printf("%d",ans[0]); 16 for(int i = 1;i < cnt;i++) printf("+%d",ans[i]); 17 printf(" "); 18 } 19 20 for(int i = x;i < n;i++){ 21 ans[cnt] = a[i]; 22 dfs(i+1,sum+a[i],cnt+1); 23 while(i+1 < n && a[i] == a[i+1]) i++; 24 } 25 } 26 27 int main(){ 28 while(scanf("%d %d",&s,&n) != EOF && s){ 29 for(int i = 0;i < n;i++) scanf("%d",&a[i]); 30 printf("Sums of %d: ",s); 31 flag = 0; 32 dfs(0,0,0); 33 if(!flag) printf("NONE "); 34 } 35 return 0; 36 }
poj 2488
骑士周游列国
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int res[105][105]; int vis[105][105]; int dir[8][2] = {-2,-1, -2,1, -1,-2, -1,2, 1,-2, 1,2, 2,-1, 2,1}; int m,n; int flag; void dfs(int x,int y,int cnt){ if(flag) return; res[cnt][0] = x;res[cnt][1] = y; if(cnt == n*m) { flag = 1; for(int i = 1;i <= n*m;i++) printf("%c%d",res[i][0]-1+'A',res[i][1]); printf(" "); return; } for(int i = 0;i < 8;i++){ int xx = x+ dir[i][0]; int yy = y+ dir[i][1]; if(!vis[xx][yy] && xx>0 && xx <= n && yy > 0 && yy <= m){ vis[xx][yy] = 1; dfs(xx,yy,cnt+1); vis[xx][yy] = 0; } } } int main(){ int T; scanf("%d",&T); int kase = 0; while(T--){ scanf("%d %d",&m,&n); memset(vis,0,sizeof(vis)); vis[1][1] = 1; printf("Scenario #%d: ",++kase); flag = 0; dfs(1,1,1); if(!flag) printf("impossible "); if(T) printf(" "); } return 0; }
poj 3009
这道题自己没有写出来
后来和题解比较了一下，这几个地方写错了
把障碍物击碎后的状态，先改成0搜一遍，等搜完还要再改回1，是为了搜到以后的所有情况
找到起点后，把起点改成0
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int INF = 1000000000; 9 int g[25][25]; 10 int vis[25][25]; 11 int n,m; 12 int sx,sy,ex,ey; 13 int flag,ans; 14 int dir[4][2] = {1,0,0,1,-1,0,0,-1}; 15 16 int in(int x,int y){ 17 if(x >= 1 && x <= m && y >=1 && y <= n) return 1; 18 return 0; 19 } 20 21 void dfs(int x,int y,int step){ 22 if(step > 10) return; 23 24 int i,xx,yy; 25 for( i = 0;i < 4;i++){ 26 xx = x+dir[i][0]; 27 yy = y+dir[i][1]; 28 29 if(step < ans && in(xx,yy) && g[xx][yy] != 1){ 30 while(g[xx][yy] == 0 && in(xx,yy)){ 31 xx +=dir[i][0]; 32 yy +=dir[i][1]; 33 } 34 int xx2 = xx-dir[i][0]; 35 int yy2 = yy-dir[i][1]; 36 if(in(xx,yy)){ 37 if(g[xx][yy] == 3){ 38 ans = step+1; 39 return; 40 } 41 if(g[xx][yy] == 1){ 42 g[xx][yy] = 0;//碰到障碍物，将障碍物清除 43 dfs(xx2,yy2,step+1); 44 g[xx][yy] = 1;//这一次dfs完后改回来是为了遍历完以后的所有情况 45 } 46 } 47 } 48 } 49 } 50 51 int main(){ 52 while(scanf("%d %d",&n,&m) != EOF && (m+n)){//输入的是n行m列 53 memset(g,0,sizeof(g)); 54 for(int i = 1;i <= m;i++){ 55 for(int j = 1;j <= n;j++){ 56 scanf("%d",&g[i][j]); 57 if(g[i][j] == 2) { 58 sx = i,sy = j; 59 g[i][j] = 0; 60 } 61 } 62 } 63 64 ans = 12; 65 dfs(sx,sy,0); 66 if(ans > 10) printf("-1 "); 67 else printf("%d ",ans); 68 } 69 return 0; 70 }
poj 1321
给出棋盘，旗子摆放不能同行或者同列，问放k个有多少种办法
感觉和poj 1564有一点点像，搜行，看对应的列是否被占用
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 typedef long long LL; 8 char g[15][15]; 9 int vis[15]; 10 LL ans; 11 int n,k; 12 13 void dfs(int x,int cnt){ 14 if(cnt == k){ 15 ans++; 16 // printf("ans = %d ",ans); 17 return; 18 } 19 for(int i = x;i <= n;i++){ 20 for(int j = 1;j <= n;j++){ 21 if(g[i][j] == '#' && !vis[j]){ 22 vis[j] = 1; 23 dfs(i+1,cnt+1); 24 vis[j] = 0; 25 } 26 } 27 } 28 } 29 30 int main(){ 31 while(scanf("%d %d",&n,&k) != EOF){ 32 if(n == -1 && k == -1) break; 33 for(int i = 1;i <= n;i++) 34 for(int j = 1;j <= n;j++) cin>>g[i][j]; 35 36 ans = 0; 37 dfs(1,0); 38 printf("%I64d ",ans); 39 } 40 return 0; 41 }
uva 11396
给出一个无向图，再给出一个爪形，问这个无向图是否能够分成一个个爪形的单元
判断是否是二分图
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 505; 9 vector<int> g[maxn]; 10 int color[maxn]; 11 int n; 12 13 bool bipartite(int u){ 14 for(int i = 0;i < g[u].size();i++){ 15 int v = g[u][i]; 16 if(color[v] == color[u]) return false; 17 if(!color[v]) { 18 color[v] = 3-color[u]; 19 if(!bipartite(v)) return false; 20 } 21 } 22 return true; 23 } 24 25 int main(){ 26 int u,v; 27 while(scanf("%d",&n) != EOF && n){ 28 for(int i = 1;i <= n;i++) g[i].clear(); 29 while(scanf("%d %d",&u,&v) != EOF){ 30 if(u == 0 && v == 0) break; 31 g[u].push_back(v);g[v].push_back(u); 32 } 33 memset(color,0,sizeof(color)); 34 color[1] = 1; 35 if(bipartite(1)) puts("YES"); 36 else puts("NO"); 37 } 38 return 0; 39 }