一、双指针

例题

日志统计

题目信息

思路

题解

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
#define maxsize 100100
using namespace std;int n,d,k;
int cnt[maxsize];
pair <int,int> num[maxsize];
bool flag[maxsize];signed main()
{ios::sync_with_stdio();cin.tie(0);cout.tie(0);cin>>n>>d>>k;for(int i=0;i<n;i++){cin>>num[i].first>>num[i].second;}sort(num,num+n);for(int i=0,j=0;j<n;j++){cnt[num[j].second]++;while(num[j].first-num[i].first>=d){cnt[num[i].second]--;i++;}if(cnt[num[j].second]>=k) flag[num[j].second]=true;}for(int i=0;i<=100000;i++){if(flag[i]) cout<<i<<endl;}return 0;
}

二、BFS

适合用于查找一条合法的路径或最小步数

例题

1、献给阿尔吉侬的花束

题目信息

思路

题解

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define maxsize 210
using namespace std;int t,r,c;
char maps[maxsize][maxsize];
int dis[maxsize][maxsize];
pair <int,int> weizhi;int bfs(pair<int,int> start,pair<int,int> end)
{queue<pair<int,int>> q;memset(dis,-1,sizeof(dis));dis[start.first][start.second]=0;q.push(start);int dx[4]={-1,0,1,0};int dy[4]={0,1,0,-1};while(!q.empty()){auto tmp=q.front();q.pop();for(int i=0;i<4;i++){int x=tmp.first+dx[i];int y=tmp.second+dy[i];//越界的情况if(x<0 || y<0 || x>=r || y>=c) continue;//障碍物if(maps[x][y]=='#') continue;//之前已经遍历过if(dis[x][y] != -1 ) continue;dis[x][y]=dis[tmp.first][tmp.second]+1;if(end==make_pair(x,y)) return dis[x][y];q.push(make_pair(x,y));}}return -1;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);pair<int,int> start,end;cin>>t;for(int i=0;i<t;i++){cin>>r>>c;for(int j=0;j<r;j++){for(int k=0;k<c;k++){cin>>maps[j][k];if(maps[j][k]=='S'){start.first=j;start.second=k;}if(maps[j][k]=='E'){end.first=j;end.second=k;} }     }int distance=bfs(start,end);if(distance==-1){cout<<"oop!"<<endl;}else{cout<<distance<<endl;}}return 0;
}

2、红与黑

题目信息

思路

题解

#include <bits/stdc++.h>
#define int long long
#define maxsize 30
using namespace std;int h,w;
char maps[maxsize][maxsize];
bool flag[maxsize][maxsize];
queue<pair<int,int>> q;int bfs(pair<int,int> start)
{int count=0;int dx[4]={-1,0,1,0};int dy[4]={0,1,0,-1};q.push(start);flag[start.first][start.second]=true;count++;while(!q.empty()){auto tmp=q.front();q.pop();for(int i=0;i<4;i++){int x=tmp.first+dx[i];int y=tmp.second+dy[i];if(x<0 || y<0 || x>=h || y>=w) continue;if(maps[x][y]=='#') continue;if(flag[x][y]==true) continue;q.push(make_pair(x,y));flag[x][y]=true;count++;}}return count;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);pair<int,int> start;while(true){cin>>w>>h;if(w==0&&h==0) break;for(int i=0;i<h;i++){for(int j=0;j<w;j++){cin>>maps[i][j];if(maps[i][j]=='@'){start.first=i;start.second=j;}}}int cnt=bfs(start);cout<<cnt<<endl;//恢复现场memset(maps,0,sizeof(maps));memset(flag,false,sizeof(flag));}return 0;
}

图论

例题

交换瓶子

题目信息

思路

题解

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define maxsize 10010
using namespace std;int n;
int a[maxsize];
bool num[maxsize];signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>n;int cnt=0;for(int i=1;i<=n;i++) cin>>a[i];for(int i=1;i<=n;i++){if(!num[i]){cnt++;for(int j=i;!num[j];j=a[j]){num[j]=true;}}}cout<<n-cnt<<endl;return 0;
}