校内模拟赛解题报告（考后总结）———2019.11.06

T1 题目链接

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

long long n,L,R;
long long f[1000];

void print(long long x,long long l,long long r){
    if(x==0){
        printf("0");
        return;
    }
    if(x==1){
        printf("1");
        return;
    }
    if(l>=f[x-2]) print(x-1,l-f[x-2],r-f[x-2]);
    else if(r<f[x-2]) print(x-2,l,r);
    else print(x-2,l,f[x-2]-1),print(x-1,0,r-f[x-2]);
}

int main(){
    std::cin>>n>>L>>R;
    if(n==0) {printf("0\n");return 0;}
    if(n==1) {printf("1\n");return 0;}
    f[0]=1;f[1]=1;
    int i;
    for(i=2;i<=n;i++){
        f[i]=f[i-1]+f[i-2];
        if(f[i]>R) break;
    }
    if(L==0){
        if(n%2==0){
            if(R==0) {printf("0\n");return 0;}
            else printf("01");
        }else{
            if(R==0) {printf("1\n");return 0;}
            else printf("10");
        }
        L=2;
    }if(L==1){
        if(n%2==0) printf("1");
        else printf("0");
        L=2;
    }
    print(i,L,R);
    std::cout<<std::endl;
    return 0;
}

View Code

T2题目链接

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>

const int inf=2147483647;
const int N=200100;
int fa[N],dep[N],mindep[N];
int e[N*2],next[N*2],point[N],t;
int n,k;

void merge(int u,int v){
    e[++t]=v;
    next[t]=point[u];
    point[u]=t;
}
void build(int x){
    int ok=0;
    for(int i=point[x];i;i=next[i]){
        int v=e[i];
        if(v==fa[x]) continue;
        fa[v]=x;
        dep[v]=dep[x]+1;
        ok=1;
        build(v);
        mindep[x]=std::min(mindep[x],mindep[v]+1);
    }
    if(!ok){
        mindep[x]=0;
        return;
    }
}
int dfs(int x){
    if(dep[x]>=mindep[x]) return 1;
    int ret=0;
    for(int i=point[x];i;i=next[i]){
        int v=e[i];
        if(v==fa[x]) continue;
        ret+=dfs(v);
    }
    return ret;
}
int main(){
    std::cin>>n>>k;
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        merge(u,v);
        merge(v,u);
    }
    for(int i=1;i<=n;i++) mindep[i]=inf;
    dep[k]=0;
    build(k);
    std::cout<<dfs(k)<<std::endl;
    return 0;
}

View Code

T3题目链接

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
#define inf 100000000
int n,m,k,l,a,b,c,ins[maxn],scc[maxn],cnt=0,num=0,dfn[maxn],low[maxn],in[maxn],dmi[maxn],dmx[maxn],visit[maxn]
,sv[maxn],val[maxn];
struct edge{
    int from,to;
};vector<edge>edges;vector<int>g[maxn];vector<edge>mp[maxn];vector<edge>mp2[maxn];
void add_edge(int f,int t){edges.push_back({f,t});g[f].push_back(k++);
}stack<int>s;
void tarjan(int u){
    low[u]=dfn[u]=++num;ins[u]=1;s.push(u);
    for(int i=0;i<g[u].size();i++){
        edge &e=edges[g[u][i]];
        if(!dfn[e.to]){tarjan(e.to);
            low[u]=min(low[u],low[e.to]);
        }else if(ins[e.to])
            low[u]=min(low[u],dfn[e.to]);
    }if(low[u]==dfn[u]){
            int y;cnt++;
            do{y=s.top();ins[y]=0;scc[y]=cnt;
                val[cnt]+=sv[y];s.pop();
            }while(u!=y);
    }
}int main(){
    cin>>n>>m;for(int i=1;i<=n;i++)sv[i]=1;
    for(int i=1;i<=m;i++)cin>>a>>b,add_edge(a,b);
    for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);
    for(int i=0;i<edges.size();i++){
        int f=edges[i].from,t=edges[i].to;
        if(scc[f]!=scc[t]){mp[scc[f]].push_back({scc[f],scc[t]});
            mp2[scc[t]].push_back({scc[t],scc[f]});
        }
    }queue<int>q;memset(dmx,0,sizeof(dmx));q.push(scc[1]);dmx[scc[1]]=val[scc[1]];visit[scc[1]]=1;
    while(!q.empty()){
        int u=q.front();q.pop();visit[u]=0;
        for(int i=0;i<mp[u].size();i++){
            edge &e=mp[u][i];
            if(dmx[e.to]<dmx[u]+val[e.to]){
                dmx[e.to]=dmx[u]+val[e.to];
                if(visit[e.to]==0){
                    q.push(e.to);
                }
            }
        }
    }for(int i=1;i<=n;i++)dmi[i]=0;
    q.push(scc[1]);dmi[scc[1]]=val[scc[1]];visit[scc[1]]=1;
    memset(visit,0,sizeof(visit));
    while(!q.empty()){
        int u=q.front();q.pop();visit[u]=0;
        for(int i=0;i<mp2[u].size();i++){
            edge &e=mp2[u][i];
            if(dmi[e.to]<dmi[u]+val[e.to]){
                dmi[e.to]=dmi[u]+val[e.to];
                if(visit[e.to]==0){
                    q.push(e.to);
                }
            }
        }
    }
    int ans=-inf;
    for(int i=0;i<edges.size();i++){
        edge &e=edges[i];
        if(dmi[scc[e.to]]!=inf&&dmx[scc[e.from]]!=0)ans=max(dmi[scc[e.to]]+dmx[scc[e.from]]-val[scc[1]],ans);
        if(dmi[scc[e.from]]!=inf&&dmx[scc[e.to]]!=0)ans=max(dmi[scc[e.from]]+dmx[scc[e.to]]-val[scc[1]],ans);
    }cout<<ans<<endl;
    return 0;
}

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标签：06,考后,int,printf,long,maxn,2019.11,return,include
来源： https://www.cnblogs.com/ydclyq/p/11818267.html