针对一类特殊图求最短路，如果边权只有01则可以使用双端队列代替堆，将最短路的时间复杂度从 \(O(nlogn)\) 降为 \(O(n)\) 。原理：每次所走边边权为0则放队首，边权为1则放队尾。
题目1
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define inf 0x3f3f3f3f
#define db double
#define il inline
#define x first
#define y second
#define endl '\n'
const int N=505;
const int mod=998244353;
char a[N][N];
int n,m;
int fx[4][2]={0,1,0,-1,1,0,-1,0};
int vis[N][N];
int safe(int x,int y) { return (x>=1)&&(x<=n)&&(y>=1)&&(y<=m); }
struct node { int x,y,w; };
void solve()
{while(cin>>n>>m&&n&&m){memset(vis,0,sizeof vis);for(int i=1;i<=n;i++){string p;cin>>p;for(int j=1;j<=m;j++) a[i][j]=p[j-1];}int x1,y1,x2,y2;cin>>x1>>y1>>x2>>y2;deque<node>q;q.push_back({x1+1,y1+1,0});vis[x1+1][y1+1]=1;while(q.size()){auto [x,y,w]=q.front();q.pop_front();if(x==x2+1&&y==y2+1) {cout<<w<<endl;break;}for(int i=0;i<4;i++){int nx=x+fx[i][0],ny=y+fx[i][1];if(safe(nx,ny)&&vis[nx][ny]==0){vis[nx][ny]=1;if(a[x][y]==a[nx][ny]) q.push_front({nx,ny,w});else q.push_back({nx,ny,w+1});}}}}
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T=1;//cin>>T;while(T--){solve();}return 0;
}
题目2
#include <bits/stdc++.h>
using namespace std;
//#define ll long long
#define pii pair<int,int>
#define inf 0x3f3f3f3f
#define db double
#define il inline
#define endl '\n'
const int N=2e3+5;
const int mod=998244353;
char a[N][N];
int n,m;
int fx[4][2]={0,1,0,-1,1,0,-1,0};
int vis[N][N];
int safe(int x,int y) { return (x>=1)&&(x<=n)&&(y>=1)&&(y<=m); }
struct node { int x,y,l,r; };
void solve()
{cin>>n>>m;int x,y;cin>>x>>y;int ll,rr;cin>>ll>>rr;for(int i=1;i<=n;i++){string p;cin>>p;for(int j=1;j<=m;j++) a[i][j]=p[j-1];}deque<node>q;q.push_back({x,y,0,0});vis[x][y]=1;int ans=0;while(q.size()){auto [x,y,l,r]=q.front();q.pop_front();if(l<=ll&&r<=rr) ans++;for(int i=0;i<4;i++){int nx=x+fx[i][0],ny=y+fx[i][1];if(vis[nx][ny]) continue;if(a[nx][ny]=='*') continue;if(!safe(nx,ny)) continue;vis[nx][ny]=1;if(fx[i][1]==-1) q.push_back({nx,ny,l+1,r});else if(fx[i][1]==1) q.push_back({nx,ny,l,r+1});else q.push_front({nx,ny,l,r});}}cout<<ans<<endl;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T=1;//cin>>T;while(T--){solve();}return 0;
}
题目3
