- 窝太水了..
- 搞不动了- -
A Quicksum
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 512;
char buffer[MAX];
int main()
{
int len;
while (fgets(buffer, MAX, stdin) != NULL)
{
//printf("read %s\n", buffer);
if (buffer[0] == ‘#‘)
{
break;
}
len = strlen(buffer);
int sum = 0;
for (int i = 0; i + 1 < len; ++i)
{
sum += (buffer[i] == ‘ ‘ ? 0 : buffer[i] - ‘A‘ + 1) * (i + 1);
}
printf("%d\n", sum);
}
return 0;
}B Asteroids
- 二分图最大匹配
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 512;
int G[MAX][MAX];
int match[MAX];
bool vis[MAX];
int n, k;
int dfs(int v)
{
int t;
for (int i = 1; i <= n; ++i)
{
if (G[i][v] && !vis[i])
{
vis[i] = true;
t = match[i];
match[i] = v;
if (t == -1 || dfs(t))
{
return 1;
}
match[i] = t;
}
}
return 0;
}
int main()
{
while (~scanf(" %d %d", &n, &k))
{
memset(G, 0, sizeof(G));
int x, y;
for (int i = 0; i < k; ++i)
{
scanf(" %d %d", &x, &y);
G[x][y] = 1;
}
int ans = 0;
memset(match, -1, sizeof(match));
for (int i = 1; i <= n; ++i)
{
memset(vis, false, sizeof(vis));
ans += dfs(i);
}
printf("%d\n", ans);
}
return 0;
}C NastyHacks
#include <cstdio>
#include <cstring>
int main()
{
int T;
scanf(" %d", &T);
while (T--)
{
int r, e, c;
scanf(" %d %d %d", &r, &e, &c);
if (e - c == r)
{
puts("does not matter");
}
else if (e - c > r)
{
puts("advertise");
}
else
{
puts("do not advertise");
}
}
return 0;
}F ScoutYYFI
- operator?时忘了写返回语句…
- 早知道开-Wall了…
- 妈蛋p.p
/ \
/ \
/ \
- 首先
dp[n]=dp[n?1]?p+dp[n?2]?(1?p) - 然后我们知道，炸弹将道路分成了若干段，设炸弹位置为pos[i]，我们要做的就是分段考虑pos[j]→pos[j+1]，当然了，第一段是1→pos[0]额外算
- 在每一段[a,b]上存活的概率都是
dp[b?1]?(1?p) - 各段相乘即可。对某一段，用矩阵快速幂加速。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
struct Mat
{
double a[2][2];
Mat operator*(const Mat& B)const
{
Mat res;
for (int i = 0; i < 2; ++i)
{
for (int j = 0; j < 2; ++j)
{
res.a[i][j] = 0.0;
for (int k = 0; k < 2; ++k)
{
res.a[i][j] += a[i][k] * B.a[k][j];
}
}
}
return res;
}
void print()
{
for (int i = 0; i < 2; ++i)
{
for (int j = 0; j < 2; ++j)
{
printf("%.2f ", a[i][j]);
}
puts("");
}
}
};
Mat fast_pow(Mat A, ll n)
{
Mat res;
res.a[0][0] = res.a[1][1] = 1.0;
res.a[0][1] = res.a[1][0] = 0.0;
while (n > 0)
{
if (n & 1)
{
res = res * A;
}
A = A * A;
//res.print();
//A.print();
//printf("n = %lld\n", n);
//puts("---------");
n >>= 1;
}
return res;
}
int n;
double p;
double gao(ll from, ll to)
{
ll n = to - from + 1;
if (n == 1)
{
return 1.0;
}
else if (n <= 0)
{
return 0.0;
}
Mat mm;
mm.a[0][0] = p;
mm.a[0][1] = 1.0;
mm.a[1][0] = 1.0 - p;
mm.a[1][1] = 0.0;
Mat A;
A.a[0][0] = 1.0;
A.a[0][1] = 0.0;
A.a[1][0] = 1.0;
A.a[1][1] = 1.0;
A = A * (fast_pow(mm, n - 1));
return A.a[0][0];
}
int main()
{
vector<ll> pos;
while (~scanf(" %d %lf", &n, &p))
{
pos.clear();
ll pp;
for (int i = 0; i < n; ++i)
{
scanf(" %lld", &pp);
pos.push_back(pp);
}
sort(pos.begin(), pos.end());
if (pos[0] == 1)
{
puts("0.0000000");
continue;
}
ll left = 1;
double ans = 1.0;
for (int i = 0; i < n; ++i)
{
ans *= gao(left, pos[i] - 1);
ans *= 1.0 - p;
left = pos[i] + 1;
}
printf("%.7f\n", ans);
}
return 0;
}D HardLife
- 这啥？不知道
- 是最大密度子图
- 模板
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=400,M=5000;
const double inf=1e10,eps=1e-6;
int head[N],nc,du[N];
struct data
{
int x,y;
}po[M];
struct edge
{
int x,y,next;
double cap;
} edge[M*3];
void add(int x,int y,double cap)
{
edge[nc].x=x;
edge[nc].y=y;
edge[nc].cap=cap;
edge[nc].next=head[x];
head[x]=nc++;
edge[nc].x=y;
edge[nc].y=x;
edge[nc].cap=0;
edge[nc].next=head[y];
head[y]=nc++;
}
int num[N],h[N],S,T,n,m;
double findpath(int x,double flow)
{
if(x==T)
return flow;
double res=flow;
int pos=n-1;
for(int i=head[x]; i!=-1; i=edge[i].next)
{
int y=edge[i].y;
if(h[x]==h[y]+1&&edge[i].cap>eps)
{
double
tp=findpath(y,min(edge[i].cap,res));
res-=tp;
edge[i].cap-=tp;
edge[i^1].cap+=tp;
if(res<eps||h[S]==n)
return flow-res;
}
if(edge[i].cap>eps&&h[y]<pos)
pos=h[y];
}
if(abs(res-flow)<eps)
{
num[h[x]]--;
if(num[h[x]]==0)
{
h[S]=n;
return flow-res;
}
h[x]=pos+1;
num[h[x]]++;
}
return flow-res;
}
double Sap(double x)
{
memset(head,-1,sizeof(head));
nc=0;
for(int i=0;i<m;i++)
{
add(po[i].x,po[i].y,1.0);
add(po[i].y,po[i].x,1.0);
}
for(int i=1;i<T;i++)
{
add(S,i,(double)m);
add(i,T,(double)m+2*x-(double)du[i]);
}
double ans=0;
memset(h,0,sizeof(h));
memset(num,0,sizeof(num));
while(h[S]!=n)
ans+=findpath(S,(T-1)*m);
return (T-1.0)*m-ans;
}
bool vis[N];
int dfs(int now)
{
int cnt=1;
vis[now]=true;
for(int i=head[now];i!=-1;i=edge[i].next)
{
if(!vis[edge[i].y]&&edge[i].cap>eps)
{
cnt+=dfs(edge[i].y);
}
}
return cnt;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(du,0,sizeof(du));
S=0;T=n+1;n=T+1;
for(int i=0; i<m; i++)
{
scanf("%d%d",&po[i].x,&po[i].y);
du[po[i].x]++;du[po[i].y]++;
}
if(m==0)
{
printf("1\n1\n");
continue;
}
double ll=0,rr=(double)m,mid;
while(rr-ll>1.0/(T-1.0)/(T-1.0))
{
mid=(ll+rr)/2.0;
double tp=Sap(mid);
if(abs(tp)<eps)
rr=mid;
else
ll=mid;
}
memset(vis,false,sizeof(vis));
int ans=dfs(S)-1;
if(ans==0)
{
Sap(ll);
memset(vis,false,sizeof(vis));
ans=dfs(S)-1;
}
printf("%d\n",ans);
for(int i=1;i<T;i++)
if(vis[i])
printf("%d\n",i);
}
return 0;
}一把辛酸泪
- 孩子，加油…
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时间： 2024-10-07 21:53:09