2012提高D2算法解析-优快云博客

2012提高D2
A.同余方程。扩展欧几里得算法求逆元。写了个暴力60分。
B.借教室。线段树写错，25分。线段树本来可以拿到90分。。正解是二分答案+前缀和
C.爆0.写的暴力一堆错。。可能不在状态吧。本来暴力可以拿到30左右。。正解是二分答案+贪心+倍增。真心不好想，不好写。

A

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define pa pair<int,int>
#define N 1000010
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int a,b;
ll x,y;
int exgcd(int a,int b,ll &x,ll &y){
    if(!b){x=1;y=0;return a;}
    int res=exgcd(b,a%b,x,y);
    ll t=x;x=y;y=t-a/b*y;
}
int main(){
//  freopen("a.in","r",stdin);
    a=read();b=read();
    exgcd(a,b,x,y);
    printf("%lld\n",(x%b+b)%b);
}

B

#include <bits/stdc++.h>
using namespace std;
#define N 1000100
#define ll long long
#define inf 0x3f3f3f3f
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,m,a[N];
struct node{
    int val,st,ed;
}b[N];
ll sum[N];
bool jud(int x){
    memset(sum,0,sizeof(sum));
    for(int i=1;i<=x;++i)
        sum[b[i].st]+=b[i].val,sum[b[i].ed+1]-=b[i].val;
    for(int i=1;i<=n;++i){
        sum[i]+=sum[i-1];
        if(sum[i]>a[i]) return 0;
    }
    return 1;
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();m=read();
    for(int i=1;i<=n;++i) a[i]=read();
    for(int i=1;i<=m;++i) b[i].val=read(),b[i].st=read(),b[i].ed=read();
    int l=1,r=m,ans=0;
    if(jud(m)){puts("0");return 0;}
    while(l<=r){
        int mid=l+r>>1;
        if(jud(mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    printf("%d\n%d\n",-1,ans+1);
    return 0;
}

C

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define pa pair<ll,int>
#define N 50010
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,m,h[N],num=0,a[N],fa[N][18],son[N],Log[N],bel[N];
ll d[N][18],l=0,r=500000,ans=0;
bool mark[N],used[N];
struct edge{
    int to,next,val;
}data[N<<1];
void dfs(int x,int tp){
    bel[x]=tp;
    for(int i=h[x];i;i=data[i].next){
        int y=data[i].to;if(y==fa[x][0]) continue;
        fa[y][0]=x;d[y][0]=data[i].val;dfs(y,tp);
    }
}
void dfs1(int x){
    bool flag=1,fson=0;
    for(int i=h[x];i;i=data[i].next){
        int y=data[i].to;if(y==fa[x][0]) continue;fson=1;
        dfs1(y);if(!mark[y]) flag=0;
    }
    if(fson&&flag) mark[x]=1;
}
bool jud(ll tot){
    memset(mark,0,sizeof(mark));memset(used,0,sizeof(used));
    vector<pa>q;vector<pa>b;
    for(int i=1;i<=m;++i){
        int x=a[i];ll dx=0;
        for(int j=Log[n];j>=0;--j)
            if(fa[x][j]&&dx+d[x][j]<=tot)
                dx+=d[x][j],x=fa[x][j];
        if(x!=1) mark[x]=1;
        else q.push_back(make_pair(tot-dx,bel[a[i]]));
    }dfs1(1);if(mark[1]) return 1;
    for(int i=1;i<=son[0];++i){
        int y=son[i];if(mark[y]) continue;
        b.push_back(make_pair(d[y][0],y));
    }sort(b.begin(),b.end());sort(q.begin(),q.end());
    for(int i=0;i<q.size();++i){//到了根以后的剩余时间不够再回来，那干脆就别去根了 
        int x=q[i].second;
        if(!mark[x]&&q[i].first<d[x][0]) mark[x]=1,used[i]=1;
    }int pq=0,pb=0;
    while(pq<q.size()&&pb<b.size()){//贪心地分配军队 
        if(used[pq]){pq++;continue;}
        if(mark[b[pb].second]){pb++;continue;}
        if(q[pq].first<b[pb].first) pq++;
        else pq++,pb++;
    }while(pb<b.size()&&mark[b[pb].second]) pb++;
    if(pb==b.size()) return 1;
    else return 0;
}
int main(){
//  freopen("blockade4.in","r",stdin);
    n=read();Log[0]=-1;for(int i=1;i<=n;++i) Log[i]=Log[i>>1]+1;
    for(int i=1;i<n;++i){
        int x=read(),y=read(),val=read();
        data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].val=val;
        data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].val=val;
        if(x!=1&&y!=1) continue;
        son[++son[0]]= x==1?y:x;
        fa[son[son[0]]][0]=1;d[son[son[0]]][0]=val;
    }m=read();if(son[0]>m){puts("-1");return 0;}
    for(int i=1;i<=son[0];++i) dfs(son[i],son[i]);
    for(int i=1;i<=Log[n];++i)
        for(int j=1;j<=n;++j)
            fa[j][i]=fa[fa[j][i-1]][i-1],d[j][i]=d[fa[j][i-1]][i-1]+d[j][i-1];
    for(int i=1;i<=m;++i) a[i]=read();
    while(l<=r){
        ll mid=l+r>>1;
        if(jud(mid)) r=mid-1,ans=mid;
        else l=mid+1;
    }
    printf("%lld\n",ans);
    return 0;
}