蓝桥练习2022

差分

#include<bits/stdc++.h>
using namespace std;
#define ll long long

ll light[500005];//灯数组 
ll cha[500005];//差分数组 
int main()
{ 
    int n,q;
    cin >> n >>q;
    for(int i = 1; i <= n; i++){
        cin >> light[i];
        cha[i]=light[i]-light[i-1];//计算差分数组 
    }
    while(q--){
        int l,r,x;
        cin >> l >> r >> x;  //差分数组常规操作 
        cha[l]+=x;            
        cha[r+1]-=x;
    }
    for(int i = 1; i <= n; i++){  //求差分数组前缀和 
        cha[i] = cha[i]+cha[i-1]; 
    }
    return 0;
 } 

最长公共子序列（不连续）

const int maxn=1e3+10;
int s1[maxn];
int s2[maxn];
int dp[maxn][maxn];
int main()
{
  // 请在此输入您的代码

  int n,m;
  cin>>n>>m;
  for(int i=1;i<=n;i++){cin>>s1[i];}
  for(int i=1;i<=m;i++) {cin>>s2[i];}

  memset(dp,0,sizeof(dp));
  for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
      //两者相等时
       if(s1[i]==s2[j]) dp[i][j]=dp[i-1][j-1]+1;
       else
       dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    }
  }
  printf("%d\n",dp[n][m]);
  return 0;
}

LIS(最长上升子序列)

  for(int i=1;i<=n;i++){
    cin>>a[i];
  }
  int len=0;
  dp[++len]=a[1];

  for(int i=2;i<=n;i++){
        if(dp[len]<a[i]) {
          dp[++len]=a[i];
        }
        else{
          int pos=upper_bound(dp+1,dp+len+1,a[i])-dp;
          dp[pos]=a[i];
          //二分在dp[]中找到比a[i]小的数最大的一个数，把a[i]插到他后面
		 //此时q[]存的就是上升子序列
        }
  }

Floyd

#include <iostream>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
const long long INF=0x3f3f3f3f3f3f3f;
long long mp[500][500];
int n,m,q;

void init(){
 memset(mp,INF,sizeof(mp));  
 for(int i=1;i<=n;i++){
    mp[i][i]=0;
  }
}
void Floyd()
{
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
            }
        }
    }
}

int main()
{
  // 请在此输入您的代码
  cin>>n>>m>>q;
  init();
  for(int i=1;i<=m;i++){
      long long u,v,w;
        scanf("%lld%lld%lld",&u,&v,&w);
        if(mp[u][v]>w)
            mp[u][v]=mp[v][u]=w;
    }
  Floyd();
  while(q--){
    long long u,v;
    cin>>u>>v;
    if(mp[u][v]<INF)
      printf("%lld\n",mp[u][v]);
    else
      printf("-1\n");
  }

  return 0;

}

最小生成树(Kruskal)

#include<cstdio>
#include<iostream>
#include<algorithm>
#define N 100010
using namespace std;
int n, m, ans = 0, cnt = 0;
int head[N], fa[N];
struct edge{
	int u, v, w;
}e[N << 1];
inline int read(){
	int s = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') ch = getchar();
	while(ch >= '0' && ch <= '9'){s = s * 10 + ch - '0'; ch = getchar();}
	return s;
}
bool cmp(edge a, edge b){return a.w < b.w;}
int find(int x){	//!!!非常重要,并查集模板 
	if(x != fa[x]) fa[x] = find(fa[x]);
	return fa[x];
}
void Kruskal(){	//Kruskal模板 
	sort(e + 1, e + m + 1, cmp);
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 1; i <= m; i++){
		int x = find(e[i].u), y = find(e[i].v);
		if(x == y) continue;	//如果两条边已经联通那么跳过 
		ans += e[i].w;
		fa[x] = y;	//将两个点合并 
		if(++cnt == n - 1) break; //当n - 1条边时结束 
	}
}
int main(){
	n = read(); m = read();
	for(int i = 1; i <= m; i++){
		int x, y, z; 
		x = read(); y = read(); z = read();
		e[i].u = x; e[i].v = y; e[i].w = z;
	}
	Kruskal();
	printf("%d\n", ans);
	return 0;
}

#include <iostream>
#include <cstdio>
#define inf 0x7fffffff
using namespace std;
int n,m,cnt,head[5005];
struct node{int next,to,w;}e[400005];
void addedge(int x,int y,int z){e[++cnt].to=y,e[cnt].w=z,e[cnt].next=head[x],head[x]=cnt;}
int flag[5005],dist[5005];
void prim()
{
	for(int i=1;i<=n;++i) dist[i]=inf; dist[1]=0;
	int ans=0;
	for(int i=1;i<=n;++i)
	{
		int minn=inf,k=-1;
		for(int j=1;j<=n;++j)
			if(dist[j]<minn&&!flag[j]) minn=dist[j],k=j;
		if(k==-1){puts("orz"); return;}
		flag[k]=1; ans+=dist[k];
		for(int j=head[k];j;j=e[j].next)
			if(!flag[e[j].to]&&dist[e[j].to]>e[j].w) dist[e[j].to]=e[j].w;
	}
	printf("%d\n",ans);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1,x,y,z;i<=m;++i) scanf("%d%d%d",&x,&y,&z),addedge(x,y,z),addedge(y,x,z);
	prim();
	return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int cnt,n,m,ans,head[5005],dist[5005],flag[5005];
struct node{int to,w,next;} e[400005];
void addedge(int x,int y,int z){e[++cnt].to=y; e[cnt].w=z; e[cnt].next=head[x]; head[x]=cnt;}
typedef pair <int,int> pii;
priority_queue <pii,vector<pii>,greater<pii> > q;
void prim()
{
    memset(dist,0x7f,sizeof(dist)); dist[1]=0;
    cnt=0;
    q.push(make_pair(0,1));
    while(!q.empty()&&cnt<n)
    {
        int d=q.top().first,u=q.top().second;
		q.pop();
        if(flag[u]) continue;
        ++cnt; ans+=d; flag[u]=1;
        for(int i=head[u];i;i=e[i].next)
        	if(e[i].w<dist[e[i].to]) dist[e[i].to]=e[i].w,q.push(make_pair(e[i].w,e[i].to));
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1,x,y,z;i<=m;++i) scanf("%d%d%d",&x,&y,&z),addedge(x,y,z),addedge(y,x,z);
    prim();
    if(cnt==n) printf("%d\n",ans);
    else puts("orz");
}