四道比较水的题
T1：SPFA+状压
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<queue> 5 #define INF 0x3f3f3f3f 6 using namespace std; 7 int n,m,K,id[20],dis[102][102],num,N,f[1<<14][15],ans,vis[102],mp[102][102]; 8 9 void get_dis(int num){ 10 queue<int> Q; memset(vis,0,sizeof(vis)); 11 Q.push(num); dis[num][num]=INF; vis[num]=1; 12 while (!Q.empty()){ 13 int now=Q.front(); Q.pop(); 14 for (int i=1; i<=n; i++){ 15 if (mp[now][i]!=0){ 16 dis[num][i]=max(dis[num][i],min(dis[num][now],mp[now][i])); 17 if (!vis[i]) Q.push(i),vis[i]=1; 18 } 19 } 20 } 21 } 22 23 int main(){ 24 scanf("%d%d%d", &n, &m, &K); 25 for (int i=1; i<=K; i++) scanf("%d", &id[i]); 26 memset(dis,0,sizeof(dis)); 27 for (int i=1,u,v,w; i<=m; i++) scanf("%d%d%d", &u, &v, &w),mp[v][u]=mp[u][v]=w; 28 for (int i=1; i<=n; i++) get_dis(i); 29 N=(1<<K); 30 for (int i=1; i<=K; i++) f[1<<(i-1)][i]=1; 31 for (int s=1; s<N; s++){ 32 num=0; 33 for (int i=1; i<=K; i++) if (s&(1<<(i-1))) num++; 34 for (int i=1; i<=K; i++) if (s&(1<<(i-1))){ 35 for (int j=1; j<=K; j++) if (!(s&(1<<(j-1)))) 36 if (num<=dis[id[i]][id[j]]) f[s^(1<<(j-1))][j]|=f[s][i]; 37 } 38 for (int i=1; i<=K; i++) if ((s&(1<<(i-1))) && f[s][i]){ 39 int t=min(num,dis[id[i]][1]); 40 ans=max(ans,t); 41 } 42 } 43 printf("%d ", ans); 44 return 0; 45 }
T2：裸RMQ
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 const int maxn = 25010; 7 int n,m,f[maxn][20]; 8 int main(){ 9 scanf("%d%d", &n, &m); 10 for (int i=1; i<=n; i++) scanf("%d", &f[i][0]); 11 for (int j=1; (1<<j)<=n; j++) 12 for (int i=1; i+(1<<(j-1))<=n; i++) 13 f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]); 14 for (int i=1,a,b; i<=m; i++){ 15 scanf("%d%d", &a, &b); 16 int t=log(b-a+1)/log(2); 17 printf("%d ", min(f[a][t],f[b-(1<<t)+1][t])); 18 } 19 return 0; 20 }
T3：曼哈顿距离转切比雪夫距离
1 #include<stdio.h> 2 #include<algorithm> 3 #include<string.h> 4 #define INF 1000010 5 using namespace std; 6 int n,a,b,x,y,mxx,mxy,mnx,mny; 7 int main(){ 8 scanf("%d", &n); 9 mxx=mxy=-INF; mnx=mny=INF; 10 for (int i=1; i<=n; i++){ 11 scanf("%d%d", &a, &b); 12 x=b+a; y=b-a; 13 mnx=min(mnx,x); mxx=max(mxx,x); 14 mny=min(mny,y); mxy=max(mxy,y); 15 } 16 printf("%d ", max(mxx-mnx,mxy-mny)); 17 return 0; 18 }
T4：set+最短路
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 #include<set> 5 #include<algorithm> 6 using namespace std; 7 const int maxn = 50010; 8 struct node{ 9 int x,y,id; 10 friend bool operator<(node a, node b){ 11 if (a.x==b.x) return a.y<b.y; 12 return a.x<b.x; 13 } 14 }; 15 int n,t,x[maxn],y[maxn],dis[maxn]; 16 queue<int> Q; 17 set<node> st; 18 node mk(int x, int y, int id){ 19 node a; a.x=x; a.y=y; a.id=id; return a; 20 } 21 int main(){ 22 scanf("%d%d", &n, &t); 23 for (int i=1; i<=n; i++){ 24 scanf("%d%d", &x[i], &y[i]); 25 st.insert(mk(x[i],y[i],i)); 26 } 27 dis[0]=0; x[0]=y[0]=0; 28 Q.push(0); 29 while (!Q.empty()){ 30 int now=Q.front(); Q.pop(); 31 for (int i=-2; i<=2; i++) for (int j=-2; j<=2; j++){ 32 int X=x[now]+i, Y=y[now]+j; 33 set<node>::iterator next=st.find(mk(X,Y,0)); 34 if (next!=st.end()){ 35 dis[next->id]=dis[now]+1; 36 Q.push(next->id); 37 if (Y>=t){ 38 printf("%d ", dis[next->id]); 39 return 0; 40 } 41 st.erase(next); 42 } 43 } 44 } 45 puts("-1"); 46 return 0; 47 }