BZOJ1011-1020

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1011：精确度不高，近似计算
1012：线段树
1013：每两个坐标联立一个方程，高斯消元

/**************************************************************    Problem: 1013    User: Gvolv    Language: C++    Result: Accepted    Time:0 ms    Memory:1276 kb****************************************************************/#include<cstdio>#include<iostream>#include<cmath>#include<cstdlib>#include<cstring>#include<algorithm>#define LL long long#define INF 1000000000#define eps 1e-10#define sqr(x) (x)*(x)#define pa pair<int,int>#define cyc(i,x,y) for(i=(x);i<=(y);i++)#define cy2(i,x,y) for(i=(x);i>=(y);i--)using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}    return x*f;}double a[12][12],x[12],s[12][12];int n,i,j;void gauss(){    cyc(i,1,n-1)    {        int k=i;        cyc(j,i+1,n) if(fabs(a[j][i])>fabs(a[k][i])) k=j;        cyc(j,i,n+1) swap(a[i][j],a[k][j]);        cyc(j,i+1,n)        {            double tmp=a[j][i]/a[i][i];            cyc(k,i,n+1) a[j][k]=a[i][k]*tmp-a[j][k];        }     }    x[n]=a[n][n+1]/a[n][n];    cy2(i,n-1,1)    {        double tmp=0.0;        cyc(j,i+1,n) tmp+=x[j]*a[i][j];        x[i]=(a[i][n+1]-tmp)/a[i][i];    }}int main(){    n=read();    cyc(i,1,n+1) s[i][0]=0.0;    cyc(i,1,n+1)    cyc(j,1,n)    cin>>a[i][j],s[i][j]=s[i][j-1]+sqr(a[i][j]);    cyc(i,1,n)    cyc(j,1,n)    {        a[i][j]=2*(a[i+1][j]-a[i][j]);    }    cyc(i,1,n) a[i][n+1]=s[i+1][n]-s[i][n];    gauss();    cyc(i,1,n-1) printf("%.3lf ",x[i]);printf("%.3lf\n",x[n]);    return 0;}

1014：不会->splay
1015：并查集，倒推连通块个数
1016-1018：不会
1019：dp
1020：计算几何，不准备写

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