Ural Vol1(dif>=900)

目前已AC:  2

1040.Airline Company(构造)

题目要求与每个顶点相连的所有边编号最大公约数为1，其实只要其中的两条边编号互质，所有边编号的最大公约数一定为1。我们知道相邻的数字一定互质，那么只要与一个顶点相连的所有边中有两条编号相邻，这个顶点就可以符合条件。DFS序列，对边进行编号刚好可以构造出满足要求的解，并且无解的情况是不存在的。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    , flag=;
    char ch;
    ;
    ';
    +(ch-');
    return flag?-res:res;
}
void Out(int a) {
    ) {putchar('-'); a=-a;}
    ) Out(a/);
    putchar(a%+');
}
;
//Code begin...

][], dfn[][], node[], n, tot;
vector<PII>edge;

void dfs(int x)
{
    node[x]=;
    FOR(i,,n) ) dfn[x][i]=dfn[i][x]=++tot, dfs(i);
}
int main ()
{
    int m, u, v;
    scanf("%d%d",&n,&m);
    ;
    FOR(i,,n) ) dfs(i);
    puts("YES");
    ; i<edge.size(); ++i) printf("%d ",dfn[edge[i].first][edge[i].second]);
    putchar('\n');
    ;
}

1075.Thread in a Space(高中几何)

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    , flag=;
    char ch;
    ;
    ';
    +(ch-');
    return flag?-res:res;
}
void Out(int a) {
    ) {putchar('-'); a=-a;}
    ) Out(a/);
    putchar(a%+');
}
;
//Code begin...

template<class T>
T sqr(T x)
{
    return x*x;
}

int main()
{
    double xa, ya, za, xb, yb, zb, xc, yc, zc, r;
    scanf("%lf%lf%lf", &xa, &ya, &za);
    scanf("%lf%lf%lf", &xb, &yb, &zb);
    scanf("%lf%lf%lf", &xc, &yc, &zc);
    scanf("%lf", &r);
    double ab=sqrt(sqr(xa-xb) + sqr(ya-yb) + sqr(za-zb));
    double ac=sqrt(sqr(xa-xc) + sqr(ya-yc) + sqr(za-zc));
    double bc=sqrt(sqr(xb-xc) + sqr(yb-yc) + sqr(zb-zc));
    *ac*bc));
    double a=acos(r/ac);
    double b=acos(r/bc);
    if(a+b>=c) printf("%.2lf\n", ab);
    else printf("%.2lf\n", (c-a-b)*r + sqrt(sqr(ac) - sqr(r)) + sqrt(sqr(bc) - sqr(r)));
    ;
}