(1) USACO2.1?Ordered Fractions

　　枚舉 排序即可，注意1/1

#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int N=165,L=1e5; struct fr{int a,b;fr(int q=0,int w=1):a(q),b(w){} }f[L]; int n,cnt=0; inline bool cmp(fr &x,fr &y){return (double)x.a/x.b<(double)y.a/y.b; } inline int gcd(int a,int b){return b==0?a:gcd(b,a%b); } int main(){cin>>n;for(int i=0;i<=n;i++)for(int j=i+1;j<=n;j++)if(gcd(i,j)==1)f[++cnt]=fr(i,j);sort(f+1,f+1+cnt,cmp);for(int i=1;i<=cnt;i++)printf("%d/%d\n",f[i].a,f[i].b);cout<<"1/1"; }

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?(2)?USACO1.5Number Triangles

　　基礎DP

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#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; int r,d[1001][1001],a[1001][1001]; int dp(int i,int j){if(d[i][j]>=0) return d[i][j];return d[i][j]=a[i][j]+(i==r?0:max(dp(i+1,j),dp(i+1,j+1))); } int main(){scanf("%d",&r);//cin>>r;memset(d,-1,sizeof(d));for(int i=1;i<=r;i++)for(int j=1;j<=i;j++) scanf("%d",&a[i][j]);//cin>>a[i][j];int ans=-10000000; cout<<dp (1,1); }

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(3) USACO1.2 Transformations

　　模擬

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#include <iostream> using namespace std; const int N=12; int n; char a[N][N],r[N][N],t[N][N]; bool ro90(char a[N][N]){for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(a[i][j]!=r[j][n-i+1]) return false;return true; } bool ro180(char a[N][N]){for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(a[i][j]!=r[n-i+1][n-j+1]) return false;return true; } bool ro270(char a[N][N]){for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(a[i][j]!=r[n-j+1][i]) return false;return true; } void img(char a[N][N],char t[N][N]){for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)t[i][n-j+1]=a[i][j]; } bool check(char r[N][N],char t[N][N]){for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(r[i][j]!=t[i][j]) return false;return true; }int solve(){if(ro90(a)) return 1;if(ro180(a)) return 2;if(ro270(a)) return 3;img(a,t);if(check(r,t)) return 4;if(ro90(t)) return 5;if(ro180(t)) return 5;if(ro270(t)) return 5;if(check(r,t)) return 6;return 7; } int main(int argc, const char * argv[]) {cin>>n;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>a[i][j];for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>r[i][j];cout<<solve();}

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(4) USACO1.4Mother's Milk

　　dfs，六種倒水方法，fill簡化

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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=25; int vis[N][N][N]; int l[4],ta,tb,tc; int ans[N]; inline void fill(int &a,int &b,int num){int tmp=min(l[num]-a,b);a+=tmp;b-=tmp; }void dfs(int a,int b,int c){//printf("%d %d %d\n",a,b,c);if(vis[a][b][c]) return;if(a==0) ans[c]=true;vis[a][b][c]=1;ta=a;tb=b;tc=c;//1fill(ta,tb,1);dfs(ta,tb,tc);ta=a;tb=b;tc=c;//2fill(ta,tc,1);dfs(ta,tb,tc);ta=a;tb=b;tc=c;//3fill(tb,ta,2);dfs(ta,tb,tc);ta=a;tb=b;tc=c;//4fill(tb,tc,2);dfs(ta,tb,tc);ta=a;tb=b;tc=c;//5fill(tc,ta,3);dfs(ta,tb,tc);ta=a;tb=b;tc=c;//6fill(tc,tb,3);dfs(ta,tb,tc); } int main(){cin>>l[1]>>l[2]>>l[3];dfs(0,0,l[3]);for(int i=0;i<=l[3];i++) if(ans[i]) cout<<i<<" ";}

?（5）USACO迷宮

　　裸DFS

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#include<iostream> using namespace std; int n,m,t,sx,sy,fx,fy,ans=0;int x,y; int e[8][8],vis[8][8],dx[4]={-1,0,1,0},dy[4]={0,1,0,-1}; void dfs(int x,int y){if(x<1||y<1||x>n||y>m) return;if(e[x][y]) return;if(x==fx&&y==fy) {ans++;return;}if(vis[x][y]) return;vis[x][y]=1; for(int i=0;i<4;i++) dfs(x+dx[i],y+dy[i]);vis[x][y]=0; } int main(){cin>>n>>m>>t>>sx>>sy>>fx>>fy; for(int i=0;i<t;i++) {cin>>x>>y;e[x][y]=1;}dfs(sx,sy);cout<<ans; }

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轉載于:https://www.cnblogs.com/candy99/p/5788793.html

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