[FJSC2014]折线统计

【题目描述】

　　二维平面上有n 个点(xi, yi)，现在这些点中取若干点构成一个集合S，对它们按照x 坐标排序，顺次连接，将会构成一些连续上升、下降的折线，设其数量为f(S)。如下图中，1->2,2->3,3->5,5->6（数字为下图中从左到右的点编号），将折线分为了4 部分，每部分连续上升、下降。

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　　现给定k，求满足f(S) = k 的S 集合个数。

【输入格式】

　　第一行两个整数n 和k，以下n 行每行两个数(xi, yi)表示第i 个点的坐标。

　　所有点的坐标值都在[1, 100000]内，且不存在两个点，x 坐标值相等或y 坐标值相等。

【输出格式】

　　输出满足要求的方案总数 mod 100007 的结果。

【样例输入】

5 1

5 5

3 2

4 4

2 3

1 1

【样例输出】

19

【数据范围】

　　对于 10% 的数据， n <= 10 , k = 1

　　对于 30% 的数据， n <= 1000, k = 1

　　对于 50% 的数据， n <= 1000, k <= 10

　　对于 100% 的数据， n <= 50000，0 < k <= 10

Solution

　　考场上mod少打一个0导致只有10分T T

　　10%:暴力

　　50%:考虑DP，记f[i][j][k]表示1~i中包含i的集合S，f(S)=j，k=0表示最后为上升趋势，k=1表示最后为下降趋势。显然有：

　　　　f[i][j][0]=Σ(f[k][j][0]+f[k][j-1][1])(k<i&&yi>yk)

　　　　f[i][j][1]=Σ(f[k][j][1]+f[k][j-1][0])(k<i&&yi<yk)

　　直接转移即可。

　　100%:注意到这个方程是在求前缀和，于是搞一颗树状数组维护一下即可。

 #include<cstdio>
 #include<ctime>
 #include<cstdlib>
 #include<algorithm>
 int n,m;
 struct P{int x,n;}x[],y[];
 bool operator<(const P&i,const P&j){return i.x<j.x;}
 int main()
 {
     srand(time());
     freopen("line.in","w",stdout);
     n=,m=;int i;
     for(i=;i<=n;i++)x[i].n=y[i].n=i,x[i].x=rand(),y[i].x=rand();
     std::sort(x+,x++n);std::sort(y+,y++n);
     printf("%d %d\n",n,m);
     for(i=;i<=n;i++)printf("%d %d\n",x[i].n,y[i].n);
 }

Data Maker

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 struct P{int x,y;}a[],tmp[];int n,k,ans[];
 bool operator<(const P&i,const P&j){return i.x<j.x||(i.x==j.x&&i.y<j.y);}
 void work(int x)
 {
     int i,f=,t=,fl;
     for(i=;x;x>>=,i++)
     if(x&)tmp[++t]=a[i];if(t<=)return;
     if(tmp[].y<tmp[].y)fl=;else fl=;
     for(i=;i<t;i++)
     {
         if(fl==){if(tmp[i+].y<tmp[i].y)f++,fl=;}
         else if(tmp[i].y<tmp[i+].y)f++,fl=;
     }
     ans[f]++;ans[f]%=;
 }
 int main()
 {
     scanf("%d%d",&n,&k);int i,j,d=<<n;
     for(i=;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);
     std::sort(a+,a++n);
     for(i=;i<d;i++)work(i);printf("%d\n",ans[k]);
 }

10%

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 const int mod=;
 int n,m,f[][][],ans;struct P{int x,y;}a[];
 bool operator<(const P&i,const P&j){return i.x<j.x||(i.x==j.x&&i.y<j.y);}
 int main()
 {
     scanf("%d%d",&n,&m);int i,j,k,l;
     for(i=;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);
     std::sort(a+,a++n);
     for(i=;i<=n;i++)f[][i][]=f[][i][]=;
     for(k=;k<=m;k++)
     {
         for(i=;i<=n;i++)
         for(j=;j<i;j++)
         {
             if(a[j].y<a[i].y)f[k][i][]=(f[k][i][]+f[k][j][]+f[k-][j][])%mod;
             if(a[j].y>a[i].y)f[k][i][]=(f[k][i][]+f[k][j][]+f[k-][j][])%mod;
         }
     }
     for(i=;i<=n;i++)ans=(ans+f[m][i][]+f[m][i][])%mod;printf("%d\n",ans);
 }

50%

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 const int mod=;
 int n,m,f[][][],ans,maxy;struct P{int x,y;}a[];
 bool operator<(const P&i,const P&j){return i.x<j.x||(i.x==j.x&&i.y<j.y);}
 int z0[],z1[];
 void add0(int x,int t){for(t%=mod;x<=maxy;x+=x&-x)z0[x]=(z0[x]+t)%mod;}
 void add1(int x,int t){for(x=maxy-x,t%=mod;x<=maxy;x+=x&-x)z1[x]=(z1[x]+t)%mod;}
 int gs0(int x){int f=;for(;x;x-=x&-x)f=(f+z0[x])%mod;return f;}
 int gs1(int x){int f=;for(x=maxy-x;x;x-=x&-x)f=(f+z1[x])%mod;return f;}
 int main(){
     scanf("%d%d",&n,&m);int i,j,k,l;
     for(i=;i<=n;i++){scanf("%d%d",&a[i].x,&a[i].y);if(a[i].y>maxy)maxy=a[i].y;}maxy++;
     std::sort(a+,a++n);
     for(i=;i<=n;i++)f[][i][]=f[][i][]=;
     for(k=;k<=m;k++)
     {
         memset(z0,,sizeof(z0));
         memset(z1,,sizeof(z1));
         for(i=;i<n;i++)
         {
             add0(a[i].y,f[k][i][]+f[k-][i][]);
             add1(a[i].y,f[k][i][]+f[k-][i][]);
             f[k][i+][]=gs0(a[i+].y);
             f[k][i+][]=gs1(a[i+].y);
         }
     }
     for(i=;i<=n;i++)ans=(ans+f[m][i][]+f[m][i][])%mod;printf("%d\n",ans);
 }

100%