USACO(含training section)水题合集[5/未完待续]

(1) USACO2.1 Ordered Fractions

　　枚举 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";
}

(2) USACO1.5Number Triangles

　　基础DP

#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);
}

(3) USACO1.2 Transformations

　　模拟

#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();

}

(4) USACO1.4Mother‘s Milk

　　dfs，六种倒水方法，fill简化

#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;//1
    fill(ta,tb,1);dfs(ta,tb,tc);

    ta=a;tb=b;tc=c;//2
    fill(ta,tc,1);dfs(ta,tb,tc);

    ta=a;tb=b;tc=c;//3
    fill(tb,ta,2);dfs(ta,tb,tc);

    ta=a;tb=b;tc=c;//4
    fill(tb,tc,2);dfs(ta,tb,tc);

    ta=a;tb=b;tc=c;//5
    fill(tc,ta,3);dfs(ta,tb,tc);

    ta=a;tb=b;tc=c;//6
    fill(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迷宫

　　裸DFS

#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;
}