Small W interval
**The teacher gives small W n intervals with closed ends on the x-axis. Let small W remove as few closed intervals as possible so that the remaining closed intervals do not intersect.
«Programming task:
Given n closed intervals, program and calculate the minimum number of closed intervals removed.
**
Sort according to the right point of the interval, and then add the left node to see if it can be added, greedy
using namespace std;
inline int read()
{
int f=1,res=0;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
while(isdigit(ch)){res=(res<<1)+(res<<3)+(ch&15);ch=getchar();}
return res*f;
}
int n,ans;
struct node
{
int x;
int y;
} a[10010];
inline bool cmp(node x,node y)//规则
{
if(x.y==y.y)return x.x<y.x;
return x.y<y.y;
}
int r=-0x7fffffff;
int main()
{
n=read();
for(int i=1;i<=n;i++)
{
a[i].x=read();
a[i].y=read();
if(a[i].x>a[i].y)swap(a[i].x,a[i].y);
}
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++)
{
if(a[i].x>r)r=a[i].y;
else ans++;
}
cout<<ans;
return 0;
}
Little W in seat selection
1. For a passenger, if it is the last one to choose a seat, then assuming that it intersects the itinerary of C passengers, then at least C+1 seats are required. Then enumerate each passenger to find out how many intersects with it, and just take the required maximum value.
2. It can be similar to adding value to an interval, if there is no value, then no one
#include<bits/stdc++.h>
using namespace std;
inline int read()
{
int f=1,res=0;char ch=getchar();
while(!isdigit(ch)){
if(ch=='-')f=-f;ch=getchar();}
while(isdigit(ch)){
res=(res<<1)+(res<<3)+(ch&15);ch=getchar();}
return res*f;
}
int n,sum[200010],xc[200010],sc[200010],maxx;
struct node
{
int l;
int r;
}a[200010];
inline bool cmp(node x,node y)
{
if(x.l==y.l)return x.r>y.r;
return x.l<y.l;
}
inline void problem1()
{
int ans=0;
for(int i=1;i<=maxx;i++)xc[i]+=xc[i-1];
for(int i=maxx;i>=1;i--)sc[i]+=sc[i+1];
for(int i=1;i<=n;++i)
ans=max(ans,n-(xc[a[i].l]+sc[a[i].r]));
cout<<ans<<' ';
}
inline void problem2()
{
int ans=0;
for(int i=1;i<=maxx;i++)sum[i]+=sum[i-1],ans=max(ans,sum[i]);
cout<<ans;
}
int main()
{
n=read();
for(int i=1;i<=n;i++)
{
a[i].l=read();
a[i].r=read();
sum[a[i].l]++;
sum[a[i].r]--;
xc[a[i].r]++;
sc[a[i].l]++;
maxx=max(maxx,a[i].r);
}
problem1();
problem2();
return 0;
}
Fantasy roller coaster
Find the shortest side connected with 1 to make a ring. If the distance is greater than the size of the ring, then it is nothing more than walking on the ring several times
//提供hzj的标程
#include <iostream>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>//d：与起点相连的边上最短边长度*2，即为起点直连的1个最小的环
//此题主要在于，对于长度为p的1到n的路径，若其存在从1到n的一条路径l，l的长度%d=p%d且l的长度<=p
using namespace std;
//因为只要有这样的l，p即为先在起点连接的最小的环绕((p-l)/d)圈再走l那条路径
typedef pair<int,int> PR;
#define mp make_pair
#define fi first
#define sc second
const int N=10010,M=20010,inf=0x3f3f3f3f;
int n,m,Q,p,t,Head[N],ver[M],cost[M],Next[M],tot=1,mi=100,f[N][110],d;
bool vis[N][110];//f[i][j]：由起点1开始，到i点，长度%d==j的路径中，最短的路径的长度
queue <PR> q;
void re(int &x)
{
x=0;
char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
}
inline void add(int x,int y,int z)
{
ver[++tot]=y; cost[tot]=z; Next[tot]=Head[x]; Head[x]=tot;
}
void spfa(int mo)
{
//不需要担心路径过程中更小的环，spfa过程中会一圈一圈套出来，最后求出的f[i][j]定为到i，距离%d为j的最短距离
int i,x,disx,y,disy;
memset(f,0x3f,sizeof(f));//初始化为最大值，以便后期更新
q.push(mp(1,0)); vis[1][0]=1; f[1][0]=0;//到点1的距离当然为0，%d也为0，入队
while(!q.empty())
{
x=q.front().fi; disx=q.front().sc; q.pop();//取出队首，x为当前点，disx为当前到当前点的最小距离%d的值
vis[x][disx]=0;//退队
for(i=Head[x];i;i=Next[i])
{
y=ver[i]; disy=(disx+cost[i])%mo;//y为下一点，disy为由当前点到y算出的1到y的距离%d的值
if(f[x][disx]+cost[i]<f[y][disy])//松弛操作
{
f[y][disy]=f[x][disx]+cost[i];
if(!vis[y][disy])//从1到y，路径长%d为disy的情况，没有再队列中
{
vis[y][disy]=1;
q.push(mp(y,disy));//入队
}
}
}
}
}
int main()
{
int u,v,w,i;
re(n); re(m);
for(i=1;i<=m;i++)
{
re(u); re(v); re(w);
add(u,v,w);
add(v,u,w);//无向边建双向
}
d=inf;
for(i=Head[1];i;i=Next[i]) d=min(d,cost[i]);//找到起点1的出边中最短的
re(Q);
if(d==inf)//如果起点1压根就没有出边，那肯定都不行
{
while(Q--) printf("AKTang!\n");
return 0;
}
d*=2;//把最短的出边*2，变成最小的环
spfa(d);//以最小的环为模数进行spfa
while(Q--)
{
re(p);
if(f[n][p%d]<=p) printf("AWaDa!\n");//如果可以通过绕d跑几圈，加上一个路径到n，总长为p，存在总长为p的路径
else printf("AKTang!\n");//否则不存在
}
return 0;
}