目录
10.4 2019-2020 ICPC Southwestern European Regional Programming Contest(SWERC 2019)
当时十分划水 也没什么要补的题..（怕不是要成水题集
F.Icebergs√
//计算几何模板，求多边形面积。
#include <bits/stdc++.h>
#define gc() getchar()
typedef long long LL;
const int N=1005;
const double eps=1e-10;
inline int dcmp(double x) {return fabs(x)<eps?0:x<0?-1:1;}
struct Vec
{
LL x,y;
Vec(LL x=0,LL y=0):x(x),y(y) {}
Vec operator +(const Vec &a)const {return Vec(x+a.x,y+a.y);}
Vec operator -(const Vec &a)const {return Vec(x-a.x,y-a.y);}
Vec operator *(const double a)const {return Vec(x*a,y*a);}
LL operator *(const Vec &a)const {return x*a.y-y*a.x;}
bool operator <(const Vec &a)const {return x<a.x||(x==a.x&&y<a.y);}
bool operator ==(const Vec &a)const {return !dcmp(x-a.x)&&!dcmp(y-a.y);}
};
typedef Vec Point;
struct Polygon
{
std::vector<Point> ps;
LL Area()//多边形有向面积（逆时针为正，顺时针为负）
{
LL res=0;
for(int i=1,lim=ps.size(); i+1<lim; ++i)
res+=(ps[i]-ps[0])*(ps[i+1]-ps[0]);
return abs(res);
}
}poly;
bool cmp(Point a,Point b)
{
if(atan2(a.y,a.x)!=atan2(b.y,b.x))
return atan2(a.y,a.x)<atan2(b.y,b.x);
else return a.x<b.x;
}
inline int read()
{
int now=0,f=1; char c=gc();
for(; !isdigit(c); c=='-'&&(f=-1),c=gc());
for(; isdigit(c); now=now*10+c-48,c=gc());
return now*f;
}
int main()
{
LL res=0;
for(int Ts=read(); Ts--; )
{
int n=read();
poly.ps.clear();
for(int i=1,x; i<=n; ++i) x=read(),poly.ps.push_back(Point(x,read()));
res+=poly.Area();
}
printf("%lld
",(res/2));
return 0;
}
G.Swapping Places(拓扑排序)
如果(i,j(i<j))两个位置不是朋友，那么(j)永远不可能到(i)前面，否则(i)对(j)没有限制。
所以可以考虑拓扑排序。。如果两个位置(i,j(i<j))处不是朋友，则连边(i->j)，即选完(i)后才能选(j)。
字典序最小的拓扑就是把queue换成priority_queue。
//467ms 65800KB
#include <bits/stdc++.h>
#define gc() getchar()
typedef long long LL;
const int N=1e5+5;
int A[N],las[N],dgr[N];
bool fri[205][205];
std::string str[222];
std::vector<int> to[N];
std::unordered_map<std::string,int> mps;
struct cmp
{
bool operator ()(int a,int b)
{
return A[a]>A[b];
}
};
std::priority_queue<int,std::vector<int>,cmp> q;
inline int read()
{
int now=0,f=1; char c=gc();
for(; !isdigit(c); c=='-'&&(f=-1),c=gc());
for(; isdigit(c); now=now*10+c-48,c=gc());
return now*f;
}
int main()
{
int S=read(),l=read(),n=read();
for (int i=1; i<=S; ++i) std::cin>>str[i];
std::sort(str+1,str+S+1);
for (int i=1; i<=S; ++i) mps[str[i]]=i;
std::string s1,s2;
for(int i=1,x,y; i<=l; ++i)
std::cin>>s1>>s2, x=mps[s1], y=mps[s2], fri[x][y]=1, fri[y][x]=1;
for(int i=1; i<=n; ++i) std::cin>>s1, A[i]=mps[s1];
for(int i=1; i<=n; ++i)
{
for(int j=1,x=A[i]; j<=S; ++j)
if(las[j] && !fri[x][j])
to[las[j]].push_back(i), ++dgr[i];
las[A[i]]=i;
}
for(int i=1; i<=n; ++i) if(!dgr[i]) q.push(i);
while(!q.empty())
{
int x=q.top(); q.pop();
std::cout<<str[A[x]]<<' ';
for(auto j:to[x])
if(!--dgr[j]) q.push(j);
}
return 0;
}
K.Birdwatching(BFS)√
记与起点相邻的点的集合为S。建反向图BFS。如果有一个点能从S中的两个点到达，则不行。
#include <bits/stdc++.h>
#define pr pair<int,int>
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
const int N = 1e5+10;
vector<int> e[N],vec;
int is[N], vis[N], tmp[N];
bool ban[N];
queue<pr> q;
int main()
{
int n, m, t; cin>>n>>m>>t;
for(int i=0; i<m; i++)
{
int u, v; scanf("%d%d",&u,&v);
e[v].pb(u);
}
memset(is, 0, sizeof(is));
memset(vis, 0, sizeof(vis));
for(auto &y: e[t]) q.push(mp(y,y)), is[y] = 1;
vis[t] = N;
while(!q.empty())
{
int x=q.front().first,pre=q.front().second; q.pop();
if(x!=pre) ban[x]=1;//, printf("ban x:%d pre:%d
",x,pre);
for(auto &v:e[x])
{
if(vis[v]>=2||v==pre) continue;
if(vis[v]==1 && tmp[v]==pre) continue;
else if(!vis[v]) tmp[v]=pre;
++vis[v], q.push(mp(v,pre));
}
}
for(int i=0; i<n; i++)
if(is[i] && !ban[i]) vec.pb(i);
cout<<vec.size()<<endl;
for(auto &x: vec) printf("%d
",x);
return 0;
}/*
4 4 4
1 2 2 3 3 1 3 4
5 7 0
1 2 2 3 3 1 3 0 4 0 3 4 4 2
8 11 0
1 4 1 2 4 2 2 3 3 1 3 0 2 5 5 0 5 6 6 7 7 2
12 15 0
1 4 1 2 4 2 2 3 3 1 3 0
7 2 6 7 5 6 5 0
2 8 8 9 9 10 10 11 11 5
*/