17-05-27模拟赛

T1:读入后按身高排序，枚举每个点可以与之相邻的身高的区间，dfs统计答案。

Code:

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k,cur,ans;
int a[11],pmx[11],pmn[11];
bool vis[11];
void dfs(int x,int step){
    if (step==n) {if (abs(cur-a[x])<=k) ++ans; return;}
    vis[x]=1;
    for (int i=pmn[x];i<=pmx[x];++i) if (!vis[i])dfs(i,step+1);
    vis[x]=0;
}
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;++i) scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    for (int i=1;i<=n;++i){
        pmx[i]=pmn[i]=i;
        while (pmx[i]<n&&a[pmx[i]+1]-a[i]<=k) pmx[i]++;
        while (pmn[i]>1&&a[i]-a[pmn[i]-1]<=k) pmn[i]--;
    }
    for (int i=1;i<=n;++i) {cur=a[i];dfs(i,1);}
    printf("%d",ans/n);
    return 0;
}

T2:二分答案，前缀和计算出每次拿的薪水，若大于二分的答案，则标记前一天。若被标记的天数（包括最后一天）大于领取的机会数，则return 0,反之return 1.

Code:

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define MN 100005
using namespace std;
int sum[MN],a[MN];
int n,m,l,r,ans=0;
bool check (int x){
    int cur=0,t=0;
    for (int i=0;i<n;++i){
        if (a[i+1]>x||t>=m) return 0;
        if (sum[i+1]-sum[cur]>x) cur=i,t++;    
    }
    return t<m;
}
int main()
{
    scanf("%d%d",&n,&m);sum[0]=0;
    for (int i=1;i<=n;++i) scanf("%d",&a[i]),sum[i]=sum[i-1]+a[i];
    l=0;r=sum[n];
    while (l<=r){
        int mid=(l+r)>>1;
        if (check(mid)) ans=mid,r=mid-1;else l=mid+1;
    }printf("%d",ans);
    return 0;
}

T3:

令in[i]为到达i点的总方案数，out[i]为从i出发到达n点的方案数，则通过边(u,v)的方案数为in[u]*out[v]，故搜索每一条边，找到该值的最大值即可。

st[i]表示该点是否有入边。

Code:

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define ME 50005
#define MN 5005
using namespace std;
struct ed{
    int x,y;
}a[ME];
struct edge{
    int to,next;
}e[ME<<1];
int r[MN],h[MN],num[MN],in[MN],out[MN];
bool rvis[MN],vis[MN],st[MN];
int n,m,x,y,cnt=1;
void ins(int *head,int x,int y){
    e[++cnt].to=y;e[cnt].next=head[x];head[x]=cnt;
}
int main()
{
    scanf("%d%d",&n,&m);memset(st,0,sizeof(st));
    for (int i=1;i<=m;++i){
        scanf("%d%d",&a[i].x,&a[i].y);st[a[i].y]=1;
        ins(h,a[i].x,a[i].y),ins(r,a[i].y,a[i].x);
    }
    for (int i=1;i<=n;++i){
        in[i]=st[i]?0:1;
        for (int j=r[i];j;j=e[j].next) in[i]+=in[e[j].to];
    }out[n]=1;
    for (int i=n;i;--i){
        for (int j=h[i];j;j=e[j].next) out[i]+=out[e[j].to];
    }int mx=0;
    for (int i=1;i<=m;++i) mx=max(in[a[i].x]*out[a[i].y],mx);
    printf("%d",mx);return 0;
}