noip模拟赛题解&总结(2015.10.18-2015.10.19)

Day1:T1

考场上觉得第二题比较水，于是就果断地跳进第二题的坑了，结果部分分都没来的及拿...

f[i][l]:用了i-n的数，这段子序列开始位置为l的方案数
考虑新加的一个数x，设原序列为S，
那么只会有三种放法：xSx,xxS,Sxx 
对于限制条件，为了方便先把大于转小于
考虑三种转移
对于xSx:
f[i][l]+=f[i+1][l+1]
不能有以下限制：
h[l]<h[r],h[r]<h[l]
h[x]<=h[l],h[x]<=h[r] (l+1<=x<=r-1)

对于xxS:
f[i][l]+=f[i+1][l+2]
不能有以下限制:
h[l]<h[l+1] ,h[l+1]<h[l]
h[x]<=h[l],h[x]<=h[l+1] (l+2<=x<=r)

对于Sxx:
f[i][l]+=f[i+1][l]
不能有以下限制：
h[r]<h[r-1],h[r-1]<h[r]
h[x]<=h[r-1],h[x]<=h[r] (l<=x<=r-2)

对于小于号的条件直接判
对于小于等于号的条件，开一个二维前缀和数组，
那么就相当于判矩形内是否有点，有则不成立 

#include<cstdio>#include<cstring>#include<algorithm>const int maxn=2010;using namespace std;int n,m,s[maxn][maxn];long long f[maxn][maxn];bool t[maxn][maxn];char str[5];int get(int x1,int y1,int x2,int y2){return s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1];}void prework(){memset(t,0,sizeof(t)),memset(s,0,sizeof(s));for (int i=1,a,b;i<=m;i++){scanf("%d%s%d",&a,str,&b);if (str[0]=='>') swap(a,b),str[0]='<';++s[a][b];if (str[0]=='=') ++s[b][a];else if (!str[1]) t[a][b]=1;}for (int i=1;i<=n*2;i++)for (int j=1;j<=n*2;j++)s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1];}void dp(){memset(f,0,sizeof(f));for (int i=1;i<=(n<<1)-1;i++) f[n][i]=(!t[i][i+1]&&!t[i+1][i]);for (int i=n-1;i;i--)for (int l=1,r=(n-i+1)*2;r<=n*2;l++,r++){if (!t[l][r]&&!t[r][l]&&!get(l+1,l,r-1,l)&&!get(l+1,r,r-1,r))//xSxf[i][l]+=f[i+1][l+1];if (!t[l][l+1]&&!t[l+1][l]&&!get(l+2,l,r,l+1))//xxSf[i][l]+=f[i+1][l+2];if (!t[r-1][r]&&!t[r][r-1]&&!get(l,r-1,r-2,r))//Sxxf[i][l]+=f[i+1][l];}long long ans=f[1][1]%(1ll<<60);if (ans<0) ans+=(1ll<<60);printf("%I64d\n",ans);}int main(){//freopen("dragon.in","r",stdin); freopen("dragon.out","w",stdout);while (scanf("%d%d",&n,&m)!=EOF) prework(),dp();return 0;}

T2：坑在这题上了，考场上写的是下1/8圆等差数列求和，上1/8圆分成R-R*sqrt(2)/2段暴力求和，最多要算六百万次sqrt，加上边界处理得比较坑，W+T就变成50了。

正解是考虑上1/8圆，x++时，y有可能-1,也有可能不减。

怎么避免大数开根呢，画图可知，只要判断(x+1,y-0.5)在圆内还是圆外即可

判圆内圆外就用x*x+y*y-r*r是小于0还是大于0

这样就可以通过只算乘法解决了

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>const int mod=1000000007;using namespace std;typedef long long ll;ll r,r2,lim;int ans;int main(){//freopen("circle.in","r",stdin);freopen("circle.out","w",stdout);scanf("%I64d",&r),r2=r*r,lim=(int)round(sqrt(1.0*r2/2));ans=((int)round(sqrt(r2-lim*lim)))%mod;if (ans>lim) ans+=lim,ans%=mod;else if (ans<lim) ans=0;ans+=((lim*(lim-1)/2)%mod),ans%=mod;ans+=r,ans%=mod;for (int x=0,y=r;x<lim-1;x++){//if (1ll*x*x+2ll*x+1ll*y*y-y-1ll*r*r+1>=0) y--;if (1.0*x*x+2.0*x+1.0*y*y-1.0*y+5.0/4.0-1.0*r2>0) y--;ans+=y,ans%=mod;}printf("%d\n",ans);return 0;}

T3:恶心的SPFA题，考场上这题没写部分分，又坑了...

对于没有怪兽的情况，跑双关键字SPFA即可

对于怪兽的格子，因为怪兽只有一次伤害，所以对于与怪兽相邻的点，两两连距离为2，伤害为周围其他怪兽的伤害的边即可。

对于陷阱，就做点权即可。

细节比较多，还必须写Dijkstra+Heap优化才可以过...

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define fi first#define se second#define mp(a,b) make_pair(a,b)#define PI pair<int,int>#define abs(a) (a>=0?a:-(a))const int maxn=505,maxq=maxn*maxn*10,maxm=1100000;const int dx[]={-maxn,-1,maxn,1};using namespace std;int n,m,stx,sty,edx,edy,pre[maxm],now[maxn*maxn],son[maxm],tot,st,ed;char map[maxn*maxn],s[maxn][maxn];PI dis[maxn*maxn],val[maxm];int P(int x,int y){return x*maxn+y;}PI D(int x){return mp(x/maxn,x%maxn);}void add(int a,int b,int bl,int dis){pre[++tot]=now[a],now[a]=tot,son[tot]=b,val[tot]=mp(bl,dis);}bool check(int x){int a=x/maxn,b=x%maxn;if (a>n||a<=0||b>m||b<=0||map[x]=='#') return 0;return 1;}bool near(int a,int b){int x1=D(a).fi,y1=D(a).se,x2=D(b).fi,y2=D(b).se;if (abs(x1-x2)+abs(y1-y2)<=1) return 1;return 0;}PI operator +(PI a,PI b){return mp(a.fi+b.fi,a.se+b.se);}struct node{int x,b,d;node( int x, int b, int d ){this->x=x,this->b=b,this->d=d;}bool operator <(const node & xx)const{if (b!=xx.b) return b>xx.b;return d>xx.d;}};void spfa(){priority_queue<node> q;q.push(node(st,0,0));memset(dis,63,sizeof(dis));dis[st]=mp(0,0);while (!q.empty()){int x=q.top().x;if (x==ed) break;q.pop();for (int y=now[x];y;y=pre[y]){PI tmp=dis[x]+val[y];if (islower(map[son[y]])) tmp.fi+=map[son[y]]-'a'+1;if (dis[son[y]]>tmp){dis[son[y]]=tmp;q.push(node(son[y],tmp.fi,tmp.se));}}}printf("%d %d\n",dis[ed].fi,dis[ed].se);}void prework(){for (int i=1;i<=n;i++)//not monsterfor (int j=1;j<=m;j++){int x=P(i,j);if (map[x]=='#'||isupper(map[x])) continue;for (int k1=0;k1<=3;k1++){int x1=x+dx[k1];if (!check(x1)||isupper(map[x1])) continue;int bl=0;for (int k2=0;k2<=3;k2++){int x2=x1+dx[k2];if (!check(x2)) continue;if (isupper(map[x2])) bl+=map[x2]-'A'+1;}add(x,x1,bl,1);}}for (int i=1;i<=n;i++)//monsterfor (int j=1;j<=m;j++){int x=P(i,j);if (isupper(map[x])){for (int k1=0;k1<=3;k1++){int x1=x+dx[k1];if (!check(x1)) continue;for (int k2=0;k2<=3;k2++) if (k2!=k1){int x2=x+dx[k2];if (!check(x2)) continue;int bl=0;for (int k3=0;k3<=3;k3++){int x3=x2+dx[k3];if (!check(x3)||near(x1,x3)) continue;if (isupper(map[x3])) bl+=map[x3]-'A'+1;}add(x1,x2,bl,2);}}}}}int main(){memset(map,'#',sizeof(map));scanf("%d%d",&n,&m),scanf("%d%d%d%d",&stx,&sty,&edx,&edy),st=P(stx,sty),ed=P(edx,edy);for (int i=1;i<=n;i++) scanf("%s",s[i]+1);for (int i=0;i<=n+1;i++)for (int j=0;j<=m+1;j++)map[P(i,j)]=s[i][j];prework(),spfa();return 0;}

Day2：T1树形DP。

设f[i][j]表示i号节点子树中距离为j的点的权值和，这可以递推求出。

然后对于每组询问(p,d)，就是子树中距离为d的点和+向上走i步的父亲的子树中距离为d-i的点的和 再减去走了回头路的点即可

#include<cstdio>#include<cstring>#include<algorithm>#define ls (j<<1)#define rs ((j<<1)|1)const int maxn=280000,maxt=44;typedef long long ll;using namespace std;int n,Q,a[maxn],pow[maxt],dep,dis;ll ans,f[maxn][maxt];int main(){freopen("dream.in","r",stdin);freopen("dream.out","w",stdout);scanf("%d%d",&n,&Q);pow[0]=1;for (int i=1;i<=maxt-1;i++) pow[i]=pow[i-1]<<1;for (int i=0;i<=maxt-1;i++) if (pow[i]-1==n){dep=i;break;}for (int i=1;i<=n;i++) scanf("%d",&a[i]);for (int i=pow[dep-1];i<pow[dep];i++) f[i][0]=a[i];for (int i=dep-1;i>=0;i--)for (int j=pow[i-1];j<pow[i];j++){f[j][0]=a[j];for (int k=1;k<=dep;k++)f[j][k]=f[j*2][k-1]+f[j*2+1][k-1];}//for (int i=1;i<=3;i++) printf("%d ",f[3][i]);puts("fuck");for (int i=1,p,d;i<=Q;i++){scanf("%d%d",&p,&d),ans=f[p][d];if (d==0){printf("%d\n",a[p]);continue;}for (dis=0;pow[dis]<=p;dis++);dis--;if (d>dep+dis-1){puts("0");continue;}if (dis==0){printf("%I64d\n",ans);continue;}int ed=min(dis,d);for (int j=1;j<ed;j++){ans+=f[p>>j][d-j];//printf("%d ",ans);ans-=f[p>>(j-1)][d-j-1];//printf("%d\n",ans);}if (d<=dis) ans+=a[p>>d];else{ans+=f[1][d-dis];ans-=f[p>>(dis-1)][d-dis-1];}printf("%I64d\n",ans);}return 0;}

T2：数论题，大致思路是类比辗转相除法，由f(a,b,l,r)得到f(b,a%b,x,y)就可以loga时间求出，但是推导过程想法很神...只知道推出来是对的，但是还不知道怎么想到这样推...考场上只拿了20分，不应该不拿另外40分部分分。

#include<cstdio>#include<cstring>#include<algorithm>const int inf=2e9+10;using namespace std;ll a,b,s,l,r;int ans;int gcd(int a,int b){return !b?a:gcd(b,a%b);}int f(int a,int b,int l,int r){if (!l) return 0;if (r/a>(l-1)/a) return (l-1)/a+1;int t=f(b,a%b,a-r%a,b-r%b);return (l+1ll*b*t-1)/a+1;}int getans(int a,int b,int l,int r){if (!a) return l?inf:0;int t=gcd(a,b);if (r/g==(l-1)/g) return inf;return f(a,b,l,r);}int main(){//freopen("ice.in","r",stdin);freopen("ice.out","w",stdout);scanf("%d",&T);while (T--){scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&s,&l,&r);l-=s,r-=s,a%=b;/*if (l>r) r+=b;if (r>=b) r-=b,l-=b;if (r<=b) r+=b,l+=b;*/if(l>r) r+=b; l-=s,r-=s;if(r>=b) r-=b,l-=b; if(r<0) r+=b,l+=b;if (l>=0) ans=getans(a,b,l,r);else ans=min(getans(a,b,0,r),getans(a,b,l+b,b-1));if (ans==inf) puts("Poor PPS.");else printf("%d\n",ans);}return 0;}

T3：考场上看到这题就放弃了，其实就是找黑白连通块数目就可以区分10个人了，剩下两个只要比较黑白区域的比例即可，代码也不是特别长。

#include<cstdio>#include<cstring>#include<algorithm>#define mp(a,b) make_pair(a,b)#define fi first#define se second#define PI pair<int,int>using namespace std;const int maxn=905,maxq=maxn*maxn*5;const int dx[]={1,1, 1,0, 0,-1,-1,-1};const int dy[]={1,0,-1,1,-1, 1, 0,-1};const char name[][15]={"Baekhyun","Chanyeol","Chen","D.O", "Kai","Kris","Lay","Luhan", "Sehun","Suho","Tao","Xiumin"};const int bc[12]={9,5,1,1, 2,3,6,5,  5,2,2,5};const int wc[12]={2,1,3,2, 13,1,2,8, 2,8,4,2};int T,ans[2],cnt[2],cb,n,m,map[maxn][maxn],head,tail;bool bo[maxn][maxn];PI q[maxq+10];void bfs(int stx,int sty,int c){head=0,q[tail=1]=mp(stx,sty),bo[stx][sty]=1;while (head!=tail){if (++head>maxq) head=1;int x=q[head].fi,y=q[head].se;for (int i=0;i<8;i++){int nx=x+dx[i],ny=y+dy[i];if (map[nx][ny]!=c||bo[nx][ny]) continue;if (++tail>maxq) tail=1;q[tail]=mp(nx,ny),bo[nx][ny]=1;}}}int main(){scanf("%d",&T);while (T--){memset(bo,0,sizeof(bo)),memset(map,-1,sizeof(map)),cb=0,ans[0]=ans[1]=0;scanf("%d%d",&n,&m);int col=0,num,x=1,y=0;for (int i=1;i<=m;i++){scanf("%d",&num),col^=1;if (!col) cb+=num;while (num--){++y;if (y>n) y=1,x++;map[x][y]=col;}}for (int i=1;i<=n;i++) for (int j=1;j<=n;j++)if (!bo[i][j]) ans[map[i][j]]++,bfs(i,j,map[i][j]);if (ans[0]==5&&ans[1]==2){if (cb*100<18*n*n) puts("Xiumin");else puts("Sehun");continue;}for (int i=0;i<12;i++)if (ans[0]==bc[i]&&ans[1]==wc[i]){printf("%s\n",name[i]);break;}}return 0;}