FZU2140 Forever 0.5
题意：构造一些点使得满足一些条件
思路：把1个点放在（0,0）然后n-1点平均分在60度的圆弧上，这样也就是有n-1对点距离为1.0
因为是60°所以n-1里面有一对点距离是1.0
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define maxn 50
#define MAX 700010
#define mod 1000000007
#define INF 0x3f3f3f3f
#define eps 1e-5
using namespace std;
const double pi = acos(-1.0) ;
struct point
{
double x , y ;
point(){};
point( double xx , double yy )
{
x = xx ; y = yy ;
}
} ;
point operator - ( point a , point b )
{
return point( b.x-a.x , b.y-a.y ) ;
}
double cross( point a , point b )
{
return a.x*b.y - a.y*b.x ;
}
int cmp( point a , point b )
{
return a.y < b.y || a.y == b.y && a.x < b.x ;
}
double get_erea(point *p,int n)
{
double ans=0;
for(int i = 1 ; i < n-1 ;i++)
{
ans += cross(p[i]-p[0],p[i+1]-p[0]) ;
}
return ans/2;
}
double len(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
point Rotate(point a,double rad)
{
return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad)) ;
}
void out(point *p,int n )
{
for( int i = 0 ; i < n ;i++)
printf("%.6lf %.6lf\n",p[i].x,p[i].y) ;
}
bool check(point *p,int n )
{
int ans=0,i,j;
for( i = 0 ; i < n ;i++)
for( j = i+1 ; j< n;j++)
{
double u = len(p[i],p[j]) ;
if(fabs(u-1.0) <eps)ans++;
else if(u>1.0) return false;
}
if(ans==n) return true;
return false;
}
int main()
{
int T,case1=0,n ;
int i,j,k,u,v,ans;
point p[110],aa ;
double rad;
cin >> T ;
while(T--)
{
scanf("%d",&n) ;
if(n<=3)
{
puts("No");
continue ;
}
p[0].x=0;
p[0].y=0;
p[1].x=1;
p[1].y=0;
aa=p[1] ;
rad=pi/3;
rad /= (n-2);
for(i = 2 ; i < n ;i++)
{
aa=Rotate(aa,rad) ;
p[i]=aa;
}
double ans=get_erea(p,n) ;
// printf("%.6lf\n",ans) ;
// cout<<check(p,n) << endl;
if(ans+eps>=0.5&&ans<=0.75)
{
puts("Yes") ;
out(p,n) ;
}
else puts("No") ;
}
return 0;
}
FZU2141 Sub-Bipartite Graph
题意：取出原图的一个子图，这个图是二分图，有至少m/2条边
思路：设二分图的两个部分是A，B，开始所有点在B里面。每次寻找最多度数（去掉A里面的点）的一个点放到A里面，然后判断是否满足要求
没有就继续上面过程。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define u64 unsigned long long
#define maxn 100010
#define mod 1000000007
#define INF 1000010
#define eps 1e-5
using namespace std;
vector<int>qe[110],ans ;
int in[110];
bool vi[110] ,hehe[110];
int find(int n)
{
memset(in,0,sizeof(in)) ;
for(int i = 1 ; i <= n ;i++)if(!hehe[i])
{
for(int j = 0 ; j < qe[i].size();j++)
{
int v = qe[i][j] ;
if(!hehe[v])in[i]++ ;
}
}
int id=INF,Max=0;
for( int i = 1 ; i <= n;i++)if(!hehe[i])
{
if(Max<in[i])
{
Max=in[i] ;
id=i;
}
}
return id;
}
bool check(int n,int m)
{
int cnt=0;
for(int i = 1 ; i <= n ;i++)if(hehe[i])
{
for( int j = 0 ; j < qe[i].size();j++)
{
int v = qe[i][j] ;
if(!hehe[v])cnt++;
}
}
if(cnt>=m/2) return true;
return false;
}
int main()
{
int i,j,n,m,k;
int u,v,len ,cnt ;
int T;
cin >> T ;
while(T--)
{
scanf("%d%d",&n,&m) ;
ans.clear();
for( i = 1 ; i <= n ;i++)
{
hehe[i]=false;
qe[i].clear();
}
for( i = 1 ; i <= m;i++)
{
scanf("%d%d",&u,&v) ;
qe[u].push_back(v) ;
qe[v].push_back(u) ;
}
len=0;
while(true)
{
k=find(n) ;
if(k==INF) break ;
hehe[k]=true;
ans.push_back(k) ;
if(check(n,m)) break ;
}
cout << ans.size() ;
for( i = 0 ; i < ans.size();i++)
cout << " " << ans[i] ;
puts("") ;
cout << n-ans.size() ;
for( i = 1 ; i <= n ;i++)
{
if(!hehe[i]){
printf(" %d",i) ;
}
}
puts("") ;
}
return 0 ;
}
FZU2143 Board Game
题意：不好说
思路：棋盘模型，用最小费用流做。
（a-b）^2 = a*a-2*a*b+b*b，b是常数，忽略。
对于偶数的格子，源点连向它k条边，第i条边的流量为1，花费是 i-1增加到i的代价
也就是 2*i-1-2*b
对于奇数的格子也是一样，不过是它连向汇点
然后奇数格子向相邻的格子连边，代价 0，流量INF
在跑最小费用流的时候，如果dis[t] >= 0 那么就可以退出了，
因为如果加上，那么总的代价就会增加
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define u64 unsigned long long
#define maxn 100010
#define mod 1000000007
#define INF 1000010
#define eps 1e-5
using namespace std;
struct node
{
int u,v,cap,flow,cost,next;
}edge[maxn];
int inf=INF,dis[1010],pre[10010],top;
int head[10010];
bool vis[10010] ;
void add(int u,int v,int cap,int flow,int cost)
{
edge[top].u = u ;
edge[top].v = v ;
edge[top].cap = cap ;
edge[top].flow = flow ;
edge[top].cost = cost ;
edge[top].next = head[u] ;
head[u]=top++ ;
}
void Unit(int u,int v,int cap,int cost )
{
//cout<<u<<" "<<v<<endl;
add(u,v,cap,0,cost);
add(v,u,0,0,-cost) ;
}
void init()
{
memset(head,-1,sizeof(head)) ;
top=0;
}
bool spfa(int s, int t)
{
queue<int>q;
// why
for (int i = 0; i < 1010; i++)
{
dis[i] = inf;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if (edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(dis[t]>=0) return false;////////././.?
if (pre[t] == -1)return false;
else return true;
}
int minCost(int s, int t)
{
int flow = 0;
int cost = 0;
while (spfa(s, t))
{
int Min = inf;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].v])
{
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].v])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
// cout<<cost<<"========"<<endl;
return cost;
}
int mat[10][10] ,map1[10][10];
int main()
{
int n,m,i,j,k,ans,s,e;
int T,case1=0,cnt;
cin >> T ;
while(T--)
{
scanf("%d%d%d",&n,&m,&k) ;
init();
ans=cnt=0;
for( i = 0 ;i < n ;i++)
for( j = 0 ; j < m ;j++){
scanf("%d",&mat[i][j]) ;
ans += mat[i][j]*mat[i][j] ;
map1[i][j]=++cnt ;
}
s = 0 ;
e = ++cnt ;
for( i = 0 ; i < n ;i++)
for( j = 0 ; j < m ;j++)
{
for( int u = 1 ; u <= k ;u++)
{
if(i%2==j%2) Unit(s,map1[i][j],1,2*u-1-2*mat[i][j]) ;
else Unit(map1[i][j],e,1,2*u-1-2*mat[i][j]) ;
}
if(i%2==j%2)
{
if(i<n-1)Unit(map1[i][j],map1[i+1][j],INF,0) ;
if(j<m-1)Unit(map1[i][j],map1[i][j+1],INF,0) ;
if(i>=1)Unit(map1[i][j],map1[i-1][j],INF,0) ;
if(j>=1)Unit(map1[i][j],map1[i][j-1],INF,0) ;
}
}
printf("Case %d: %d\n",++case1,ans+minCost(s,e)) ;
}
return 0 ;
}
FZU 2144 Shooting Game
题意：有n个物体，它出现在一些地方，有一个飞行的方向，速度为1；一个人可以一次袭击把圆里面的所以物体击落
思路：问最少多少次袭击，可以把最多的物品击落。
联立球的方程和直线方程，可以解出 进去时间和出来时间，也就是一个居间
这样就是用最少的点覆盖所有线段问题
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define u64 unsigned long long
#define maxn 100010
#define mod 1000000007
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
struct node
{
double L,R ;
bool operator<(const node&s) const
{
return R < s.R ;
}
}qe[maxn] ;
void getnumq(int& x){
char ch=getchar();
bool mk=0;
x = 0;
while(ch<48 || ch>57){
if(ch==‘-‘) mk=1;
ch=getchar();
}
while(ch>=48 && ch<=57){
x = x*10+ch-48;
ch = getchar();
}
if(mk) x=-x;
}
void getnum(double& x){
int y;
getnumq(y);
x = y*1.0;
}
int solve(int m)
{
sort(qe,qe+m) ;
int ans=0,j;
for( int i = 0 ; i < m ;i++)
{
ans++ ;
j=i+1;
while(j<m && qe[j].L <=qe[i].R )
{
j++ ;
}
i = j-1;
}
return ans;
}
int main()
{
int i,n,m,j,len,T,case1=0;
double ax,ay,az,dx,dy,dz ;
double r,a,b,c,d;
cin >> T ;
while(T--)
{
scanf("%d%lf",&n,&r) ;
len=0;
for( j = 1 ; j <= n ;j++ )
{
getnum(ax); getnum(ay); getnum(az);
getnum(dx); getnum(dy); getnum(dz);
a = dx*dx+dy*dy+dz*dz;
b = 2*(ax*dx+ay*dy+az*dz);
c = ax*ax+ay*ay+az*az-r*r;
d = b*b-4*a*c;
if(d<0)continue;
qe[len].L = (b*(-1)-sqrt(d))/(a*2);
qe[len].R = (b*(-1)+sqrt(d))/(a*2);
if(qe[len].L<0 && qe[len].R<0)continue;
len++;
}
printf("Case %d: %d %d\n",++case1,len,solve(len)) ;
}
return 0 ;
}
FZU 2148 Moon Game
题意：给出n个点，问组成凸四边形的组数
思路：数据很小，枚举。然后判断是否是凸的，
判断大概可以用凸包了......
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define maxn 50
#define MAX 700010
#define mod 1000000007
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
struct point
{
int x , y ;
point(){};
point( int xx , int yy )
{
x = xx ; y = yy ;
}
} ;
point qe[maxn] , p[maxn] ;
point operator - ( point a , point b )
{
return point( b.x-a.x , b.y-a.y ) ;
}
int cross( point a , point b )
{
return a.x*b.y - a.y*b.x ;
}
int cmp( point a , point b )
{
return a.y < b.y || a.y == b.y && a.x < b.x ;
}
int Get( int n )
{
sort(qe,qe+n,cmp) ;
int m = 1 , i , k;
if( n == 0 ) return 0 ;p[0] = qe[0] ;
if( n == 1 ) return 1 ;p[1] = qe[1] ;
if( n == 2 ) return 2 ;p[2] = qe[2] ;
for( i = 2 ; i < n ;i++)
{
while( m && cross( p[m]-p[m-1] , qe[i] - p[m-1] ) <= 0 )m-- ;
p[++m] = qe[i] ;
}
k = m ;p[++m] = qe[n-2] ;
for( i = n-3 ; i >= 0 ; i-- )
{
while( m > k && cross( p[m]-p[m-1] , qe[i]-p[m-1] ) <= 0 )m-- ;
p[++m] = qe[i] ;
}
return m ;
}
int xx[maxn],yy[maxn] ;
int main()
{
int T,case1=0,n ;
int i,j,k,u,v,ans;
cin >> T ;
while(T--)
{
scanf("%d",&n) ;
for( i = 1 ; i <= n ;i++)
scanf("%d%d",&xx[i],&yy[i]) ;
ans=0;
for( i=1 ; i<= n ;i++)
for(j=i+1;j<=n;j++)
for(u=j+1;u<=n;u++)
for(v=u+1;v<=n;v++)
{
qe[0].x=xx[i] ;qe[0].y=yy[i] ;
qe[1].x=xx[j] ;qe[1].y=yy[j] ;
qe[2].x=xx[u] ;qe[2].y=yy[u] ;
qe[3].x=xx[v] ;qe[3].y=yy[v] ;
k = Get(4) ;
//cout<<k<<endl;
if(k==4)ans++;
}
printf("Case %d: %d\n",++case1,ans) ;
}
return 0;
}
FZU 2150 Fire Game
题意:、
思路：枚举两点，然后dfs;
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<cstring>
#include<set>
#include<stack>
#include<string>
#include<ctime>
#define LL long long
#define maxn 20
#define MAX 700010
#define mod 1000000007
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
char mat[maxn][maxn] ;
int time1[maxn][maxn] ;
int n,m;
void dfs(int x,int y,int t)
{
time1[x][y]= t ;
if(x>1&&mat[x-1][y]==‘#‘&&time1[x-1][y]>t+1)
{
dfs(x-1,y,t+1);
}
if(y>1&&mat[x][y-1]==‘#‘&&time1[x][y-1]>t+1)
{
dfs(x,y-1,t+1);
}
if(x<n&&mat[x+1][y]==‘#‘&&time1[x+1][y]>t+1)
{
dfs(x+1,y,t+1);
}
if(y<m&&mat[x][y+1]==‘#‘&&time1[x][y+1]>t+1)
{
dfs(x,y+1,t+1);
}
}
int solve(int s1,int e1,int s2,int e2)
{
memset(time1,INF,sizeof(time1)) ;
dfs(s1,e1,0) ;
dfs(s2,e2,0) ;
int ans=-1 ;
for(int i = 1 ; i <= n ;i++)
for(int j = 1 ; j <= m ;j++)if(mat[i][j]==‘#‘)
{
ans=max(ans,time1[i][j]) ;
}
return ans;
}
int getans()
{
int ans=INF,t;
for(int i = 1 ; i <= n ;i++)
for(int j = 1 ; j <= m ;j++)if(mat[i][j]==‘#‘)
{
for(int k = i ; k <= n ;k++)
for(int kk = 1 ; kk <= m ;kk++)if(mat[k][kk]==‘#‘)
{
int t = solve(i,j,k,kk) ;
if(t==-1) t = INF ;
ans=min(ans,t) ;
}
}
if(ans==INF) return -1;
return ans;
}
int main()
{
int T,case1=0 ;
int i,j,k,ans;
cin >> T ;
while(T--)
{
scanf("%d%d",&n,&m) ;
for( i = 1 ; i <= n ;i++)
scanf("%s",mat[i]+1) ;
printf("Case %d: %d\n",++case1,getans()) ;
}
return 0;
}
时间： 2024-10-28 23:16:50