BZOJ 4747-4749题解 Usaco2016 Dec

2019-11-06 08:06:31

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BZOJ 4747: [Usaco2016 Dec]Counting Haybales

给出N（1≤N≤100,000）个数，和 Q（1≤Q≤100,000）个询问。每个询问包含两个整数A,B(0≤A≤B≤1,000,00 0,000）。对于每个询问，给出数值在A到B间的数有多少个（包含A与B）。暴力

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=PRe[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld/n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { / For(j,m-1) cout<<a[i][j]<<' ';/ cout<<a[i][m]<<endl; / } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){ int x=0,f=1; char ch=getchar(); while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();} return x*f;} #define MAXN (123456)int a[MAXN];int main(){// freopen("bzoj4747.in","r",stdin);// freopen(".out","w",stdout); int n=read(),q=read(); For(i,n) a[i]=read(); sort(a+1,a+1+n); For(i,q) { int p1=read(),p2=read(); int c=upper_bound(a+1,a+1+n,p1-1)-a; int d=upper_bound(a+1,a+1+n,p2)-a-1; printf("%d/n",d-c+1); } return 0;}

BZOJ 4748: [Usaco2016 Dec]Cities and States

依次给出N（1≤N≤200,000）个城市名及其州的代码（州代码保证仅有两位）。我们说，两个城市是一个“特殊对 ”，如果他们来自不同的州且A城市的州代码是B城市的城市名首两位，B城市的州代码是A城市的城市名首两位。问：在给定的信息中有多少个“特殊对”？

hash后暴力统计

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld/n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { / For(j,m-1) cout<<a[i][j]<<' ';/ cout<<a[i][m]<<endl; / } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){ int x=0,f=1; char ch=getchar(); while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();} return x*f;} map<pi ,int> h; map<pi ,int>::iterator it; int main(){// freopen("bzoj4748.in","r",stdin);// freopen(".out","w",stdout); int n=read(); For(i,n) { char s[100],s2[100]; scanf("%s%s",s,s2); int p1=(s[0]-'A')*233+s[1]-'A'; int p2=(s2[0]-'A')*233+s2[1]-'A'; if (h.count(mp(p1,p2))) h[mp(p1,p2)]++; else h[mp(p1,p2)]=1; } ll ans=0; for(it=h.begin();it!=h.end();it++) { pair<pi ,int> p=*it; pi pa=p.fi; swap(pa.fi,pa.se); if (h.count(pa)) { if (pa.fi<pa.se) ans+=h[pa]*(ll)p.se; } } printf("%lld/n",ans); return 0;}

BZOJ 4749: [Usaco2016 Dec]Moocast

每头牛手上有一台对讲机，给出N（1≤N≤200）头牛的坐标（X,Y）及其对讲机的极限传输半径P，也就是说该对讲机能将信息传送到与之距离不超过P的对讲机。幸运的是,牛可以传递消息通过其他牛，所以没有必要每头牛能直接传送到其他牛。请帮助奶牛确定：如果源于一头牛，最多能将信息传递到多少头牛？

暴力建图bfs

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld/n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { / For(j,m-1) cout<<a[i][j]<<' ';/ cout<<a[i][m]<<endl; / } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){ int x=0,f=1; char ch=getchar(); while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();} return x*f;} #define MAXN (300)vi edges[MAXN];ll n,x[MAXN],y[MAXN],p[MAXN];bool b[MAXN];int main(){// freopen("bzoj4749.in","r",stdin);// freopen(".out","w",stdout); n=read(); For(i,n) { x[i]=read(),y[i]=read(),p[i]=read(); } For(i,n) For(j,n) if (i!=j) { if ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=p[i]*p[i]) { edges[i].pb(j); } } int ans=0; For(i,n) { MEM(b) b[i]=1; int t=0; queue<int> q;q.push(i); while(!q.empty()) { t++; int now=q.front();q.pop(); Rep(j,SI(edges[now])) { if (!b[edges[now][j]]) { q.push(edges[now][j]); b[edges[now][j]]=1; } } } ans=max(ans,t); } printf("%d/n",ans); return 0;}

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