差一题AK 

题目链接

A-深度学习

做法:训练时间b为n即可，答案为n

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
int main()
{
    double n;
    cin>>n;
    printf("%f",n);
}

B-Metropolis

做法：用一个dp维护一下当前i 从哪个都市 到这个点最近，dis普通维护一下dij。中间如果出现一条边的两个端点 的dp来自不同的都市，那么就对那两个都市进行更新ans   ans[dp[u]]=min( ans[dp[u]], dis[u]+div[v]+w  )

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;
vector<pair<int,ll> >G[N];
int n,m,p,a[N];
int dp[N];//i从dp[i]都市转移过来得到的 最小值dis[i]
ll dis[N],ans[N];
struct node
{
    int u;
    ll w;
    bool operator <(const node &o) const
    {
        return w>o.w;
    }
};
void dij()
{
    for(int i=1;i<=n;++i) dis[i]=ans[a[i]]=1e18;
    priority_queue<node>que;
    for(int i=1;i<=p;++i){
        que.push({a[i],0});
        dis[a[i]]=0;
    }
    while(que.size()){
        node now=que.top();que.pop();
        //printf("u:%d\n",now.u);

        for(auto it:G[now.u]){
            if(dis[now.u]+it.second<dis[it.first]){
                dis[it.first]=dis[now.u]+it.second;
                dp[it.first]=dp[now.u];
                que.push({it.first,dis[it.first]});
            }
            else if(dp[now.u]!=dp[it.first]){

                ans[dp[now.u]]=min(ans[dp[now.u]],dis[now.u]+dis[it.first]+it.second);
                ans[dp[it.first]]=min(ans[dp[it.first]],dis[now.u]+dis[it.first]+it.second);
            }
        }
    }
}
int main()
{
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=p;++i){
        scanf("%d",&a[i]);
        dp[a[i]]=a[i];
    }
    //puts
    for(int i=1;i<=m;++i){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        G[u].push_back({v,w});
        G[v].push_back({u,w});
    }
    //puts("***");
    dij();
    for(int i=1;i<=p;++i) printf("%lld ",ans[a[i]]);
}

C-迷宫

做法:2020牛客寒假训练营的一道题，做过，其实两种走法：向右或者向下。

简单bfs一下即可。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define pi pair<int, int>
#define mk make_pair
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int N=1e3+10;
char s[N][N];
int n,m,a[N][N],vis[N][N];
int vs[N][N];
struct node
{
    int x,y,step;
    bool operator <(const node &o) const{
        return o.step<step;
    }
};
int dir[2][2]={0,1,1,0};
void bfs()
{
    priority_queue<node>que;
    que.push({1,1,0});
    vis[1][1]=1;
    while(que.size())
    {
        node now=que.top();que.pop();
        //printf("x:%d y:%d t:%d\n",now.x,now.y,now.step);
        if(now.x==n&&now.y==m){
            printf("%d\n",now.step);
            return ;
        }
        int f=0;
        for(int i=0;i<2;++i){
            int x=now.x+dir[i][0];
            int y=now.y+dir[i][1];
            if(x<1||y<1||x>n||y>m) continue;
            if(a[x][y]) continue;

            vis[x][y]=1;
            if(now.step+f<vs[x][y]){
                que.push({x,y,now.step+f});
                vs[x][y]=now.step+f;
            }

            f=1;
        }
    }
    puts("-1");
}
int main()
{
	scanf("%d%d",&n,&m);
	memset(vs,0x3f3f3f3f,sizeof(vs));

	rep(i,1,n){
        scanf("%s",s[i]+1);
        rep(j,1,m){
            a[i][j]=s[i][j]-'0';
        }
	}
	bfs();
	return 0;
}
/*
7 7
0000000
0100100
0101100
1000001
1101001
0001000
0010000
ans:2
*/

D-法法

做法：高次幂的奇偶性跟底数有关，全排列 和的 奇偶也就是  全排列中第一个数的奇偶个数。

全排列首位枚举：n个  后面n-1 全排列(n-1)!  当n-1>=2  后面的永远是偶数。

于是很显然 当n<=2 答案为1  其他为0

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
int main()
{
    int _=read();
    while(_--)
    {
        ll n=read();
        if(n<=2) puts("1");
        else puts("0");
    }
}

E-Graph Coloring I

做法：先 dfs 找 奇数环，再染色找二分图即可。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
const int N=3e5+10;
vector<int>G[N],ans;
int n,m,col[N],fa[N],dfn[N],flag;

void dfs(int u,int f)
{
    if(flag) return ;
    dfn[u]=dfn[f]+1;
    fa[u]=f;

    for(int v:G[u]){
        if(v==f) continue;
        if(dfn[v]){
            if((dfn[u]-dfn[v]+1)%2){
                flag=1;
                int now=u;
                while(now!=v){
                    //printf("now:%d v:%d\n",now,v);
                    ans.push_back(now);
                    now=fa[now];
                }
                ans.push_back(now);
                //puts("***");
                break;
            }
            continue;
        }
        dfs(v,u);
        if(flag) return ;
    }
}

void dfs2(int u,int f)
{
    for(int v:G[u]){
        //if(v==f) continue;
        if(col[v]) continue;
        col[v]=3-col[u];
        dfs2(v,u);
    }
}
int main()
{
    n=read(),m=read();
    rep(i,1,m)
    {
        int u=read(),v=read();
        G[u].push_back(v);
        G[v].push_back(u);
    }
    //dfn[1]=1;

    dfs(1,1);
    if(flag){
        printf("%d\n",ans.size());
        for(int v:ans) printf("%d ",v);
    }
    else{
        puts("0");
        col[1]=1;
        dfs2(1,1);
        rep(i,1,n) printf("%d ",col[i]-1);
    }
}
/*
4 4
1 2
2 3
3 4
4 1
*/