题目
得分应该是240,早年的题确实好做。也带点运气成分
1-4 帅到没朋友
题意: 略。
思路:
模拟即可,可以发现如果在数量>1的地方出现过,就不可能满足条件了。(我个憨憨写了并查集)。
注意: 输出要用%05d,前边补0,这个值5分。。。
时间复杂度: O(input)
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
#define mem(a,x) memset(a,x,sizeof(a))
#define fir(i,a,b) for(int i=a;i<=b;++i)
const int N = 2e5+10;
int n,m,k,T;
int cnt[N];
bool vis[N];
void solve()
{
cin>>n;
fir(i,0,N-1) cnt[i] = 1;
for(int i=1;i<=n;++i)
{
int num; cin>>num;
int x;
if(num==1)
{
cin>>x;
}
else
{
while(num--) cin>>x,cnt[x] = 0;
}
}
cin>>m;
vector<int> va;
while(m--)
{
int x; cin>>x;
if(vis[x]) continue;
vis[x] = 1;
if(cnt[x]) va.push_back(x);
}
if(va.size())
{
for(int i=0;i<va.size();++i)
{
if(i) cout<<" ";
printf("%05d",va[i]);
}
}
else cout<<"No one is handsome";
}
signed main(void)
{
// ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
solve();
return 0;
}
L3-1 天梯地图
题意: 给定n个点m条边的图,有时间和距离两种边权。求时间花费最少的路径,如果不唯一,则输出最短路(题目保证该路径唯一)。并且求距离最短的路径,如果不唯一,则输出途中经过点最少的路径。
思路: 刚开始题意看错了,没有看到句号,就写歪了。显然先求一下最短路,然后再求一下最快路,然后输出即可。反正挺麻烦的。
时间复杂度: O(mlogm)
代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 502*502;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
write(x/10);
}
putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
x = 0;char ch = getchar();ll f = 1;
while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
//1为距离、2为时间
int h[N],e[N<<1],ne[N<<1],w[N<<1],w2[N<<1],idx = 0;
int st,ed;
void add(int a,int b,int c,int d)
{
e[idx] = b,w[idx] = c,w2[idx] = d,ne[idx] = h[a],h[a] = idx++;
}
bool vis[N];
int dist[3][N];
int pre[3][N];
int cnt[N]; //顶点个数
int d[N]; //最快路中的最短
void Dij1(int S)
{
for(int i=0;i<n;++i) dist[1][i] = 0x3f3f3f3f,pre[1][i] = -1,vis[i] = false;
priority_queue<PII,vector<PII>,greater<PII> >q;
dist[1][S] = 0;
cnt[S] = 1;
q.push({0,S});
while(q.size())
{
auto tmp = q.top(); q.pop();
int u = tmp.first,dis = tmp.second;
swap(u,dis);
if(vis[u]) continue;
vis[u] = true;
for(int i=h[u];~i;i=ne[i])
{
int j = e[i];
if(dist[1][j] > dist[1][u] + w[i])
{
dist[1][j] = dist[1][u] + w[i];
cnt[j] = cnt[u] + 1;
pre[1][j] = u;
q.push({dist[1][j],j});
}
else if(dist[1][j] == dist[1][u] + w[i])
{
if(cnt[j] > cnt[u] + 1)
{
cnt[j] = cnt[u] + 1;
pre[1][j] = u;
}
}
}
}
}
void Dij2(int S)
{
for(int i=0;i<n;++i) d[i] = 0x3f3f3f3f,dist[2][i] = 0x3f3f3f3f,pre[2][i] = -1,vis[i] = false;
priority_queue<PII,vector<PII>,greater<PII> >q;
dist[2][S] = 0;
d[S] = 0;
q.push({0,S});
while(q.size())
{
auto tmp = q.top(); q.pop();
int u = tmp.first,dis = tmp.second;
swap(u,dis);
if(vis[u]) continue;
vis[u] = true;
for(int i=h[u];~i;i=ne[i])
{
int j = e[i];
if(dist[2][j] > dist[2][u] + w2[i])
{
dist[2][j] = dist[2][u] + w2[i];
pre[2][j] = u;
d[j] = d[u] + w[i];
q.push({dist[2][j],j});
}
else if(dist[2][j] == dist[2][u] + w2[i])
{
if(d[j] > d[u] + w[i])
{
d[j] = d[u] + w[i];
pre[2][j] = u;
}
}
}
}
}
void solve()
{
mem(h,-1); idx = 0;
// ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
scanf("%d%d",&n,&m);
for(int i=0;i<m;++i)
{
int x,y,op,z,z2;
scanf("%d%d%d%d%d",&x,&y,&op,&z,&z2);
add(x,y,z,z2);
if(op==0) add(y,x,z,z2);
}
scanf("%d%d",&st,&ed);
Dij1(st);
Dij2(st);
// fir(i,0,n-1) cout<<i<<":"<<dist[1][i]<<endl;
stack<int> sa,sb; //时间、距离
int x = ed;
while(x != -1)
{
sa.push(x);
x = pre[2][x];
}
x = ed; while(x != -1) {sb.push(x); x = pre[1][x];}
if(sa == sb)
{
printf("Time = %d; Distance = %d: ",dist[2][ed],dist[1][ed]);
int t = sa.top(); sa.pop();
printf("%d",t);
while(sa.size())
{
t = sa.top(); sa.pop();
printf(" => ");
printf("%d",t);
}
}
else
{
printf("Time = %d: ",dist[2][ed]);
int t;
t = sa.top(); sa.pop();
printf("%d",t);
while(sa.size())
{
t = sa.top(); sa.pop();
printf(" => ");
printf("%d",t);
}
printf("\n");
printf("Distance = %d: ",dist[1][ed]);
t = sb.top(); sb.pop();
printf("%d",t);
while(sb.size())
{
t = sb.top(); sb.pop();
printf(" => ");
printf("%d",t);
}
}
}
signed main(void)
{
T = 1;
// OldTomato; cin>>T;
// read(T);
while(T--)
{
solve();
}
return 0;
}
L3-3 长城
题意: 略。
思路:
时间复杂度:
代码: