标签：长度 void print possible min poj getc while distance
POJ 2031 Building a Space Station
猜一个结论两个球体间的最短距离是圆心间距离减去两个球的半径，如果是负数就说明相交。
然后理解了三维上两个点之间的距离怎么求
//#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<algorithm> #include<cstring> #include<map> #include<cmath> #include<vector> #define inf 2e9 #define maxnode 200000 #define ll long long #define lowbit(x) (x&(-x)) const int mod = 1e9 + 7; const int maxn = 1e5 + 10; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; using namespace std; int par[105],cnt; struct edge{ int u,v; double cost; edge(int u1=0,int v1=0,double c1=0): u(u1),v(v1),cost(c1) {} }edges[10010]; bool cmp(edge n1,edge n2){ return n1.cost<n2.cost; } int find_root(int x){ if( x==par[x] ) return x; return par[x] = find_root( par[x] ); } double x[105],y[105],z[105],r[105]; double find_dist(int i,int j){ //圆心间距离 double dis = sqrt( abs(x[i]-x[j])*abs(x[i]-x[j]) + abs(y[i]-y[j])*abs(y[i]-y[j]) + abs(z[i]-z[j])*abs(z[i]-z[j]) ); dis-=r[i]; dis-=r[j]; return dis; } int main(){ int n; while(1){ cin>>n; if(n==0) break; cnt=0; for(int i=1;i<=n;i++) par[i]=i; for(int i=1;i<=n;i++) cin>>x[i]>>y[i]>>z[i]>>r[i]; for(int i=1;i<=n;i++){ for(int j=i+1;j<=n;j++){//find distance double dist = find_dist(i,j); if( dist<=0 ){//i跟j一开始就在一个集合 int rooti = find_root(i), rootj = find_root(j); par[rooti] = rootj; } else edges[++cnt] = edge(i,j,dist); } } sort(edges+1,edges+1+cnt,cmp); //for(int i=1;i<=cnt;i++) cout<<edges[i].u<<" "<<edges[i].v<<" "<<edges[i].cost<<endl; double ans=0; for(int i=1;i<=cnt;i++){ int u = edges[i].u,v=edges[i].v; int rootu = find_root(u), rootv = find_root(v); if( rootu!=rootv ){ par[rootu] = rootv; ans+=edges[i].cost; } } printf("%.3lf\n",ans); } return 0; }
POJ 1789 Truck History
如果把每个Truck当作一个顶点的话，那任意一颗生成树都能对应上一个derive的方案，那就是求MST
//#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<algorithm> #include<cstring> #include<map> #include<cmath> #include<vector> #define inf 2e9 #define maxnode 200000 #define ll long long #define lowbit(x) (x&(-x)) const int mod = 1e9 + 7; const int maxn = 2e3 + 10; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; using namespace std; int dist[maxn],vis[maxn];//从mst集合到这个点的距离 int a[maxn][maxn]; char trucks[maxn][10]; int find_difference(int i,int j){ int ans=0; for(int k=0;k<7;k++){ if( trucks[i][k]!=trucks[j][k] ) ans++; } return ans; } int main(){ int n; while( scanf("%d",&n)){ if(n==0) break; for(int i=1;i<=n;i++) scanf("%s",trucks[i]); for(int i=1;i<=n;i++){ for(int j=i+1;j<=n;j++){ a[i][j] = a[j][i] = find_difference(i,j); } } memset(vis,0,sizeof(vis)); //mst int ans=0; for(int i=2;i<=n;i++){ dist[i] = a[1][i]; } for(int i=2;i<=n;i++){ int mind=10,index; for(int j=2;j<=n;j++){ if( vis[j]==0 && dist[j]<mind ) { mind=dist[j]; index=j; } } // ans+=dist[index]; vis[index]=1; for(int j=2;j<=n;j++){ if( j!=index && dist[j]>a[index][j] ) dist[j] = a[index][j]; } } printf("The highest possible quality is 1/%d.\n",ans); } return 0; }
POJ 2349 Arctic Network
知道结论：所有最小生成树边按边的权值排序后，序列一样
找第k大的边就行了
//#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<algorithm> #include<cstring> #include<map> #include<cmath> #include<vector> #define inf 2e9 #define maxnode 200000 #define ll long long #define lowbit(x) (x&(-x)) const int mod = 1e9 + 7; const int maxn = 5e2 + 10; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; using namespace std; int a[maxn][maxn],dist[maxn]; int x[maxn],y[maxn],vis[maxn]; int find_dist(int i,int j){ return abs(x[i]-x[j])*abs(x[i]-x[j])+abs(y[i]-y[j])*abs(y[i]-y[j]); } int main(){ int t; scanf("%d",&t); while( t-- ){ int s,n; scanf("%d%d",&s,&n); for(int i=1;i<=n;i++) scanf("%d%d",x+i,y+i); for(int i=1;i<=n;i++){ for(int j=i+1;j<=n;j++){ a[i][j] = a[j][i] = find_dist(i,j); // cout<<i<<" "<<j<<" "<<a[i][j]<<endl; } } memset(vis,0,sizeof(vis)); //mst for(int i=2;i<=n;i++) dist[i] = a[1][i]; vector<int> edge; edge.clear(); for(int i=2;i<=n;i++){ int mind=inf,index; for(int j=2;j<=n;j++){ if( vis[j]==0 && dist[j]<mind ) { mind=dist[j]; index=j; } } // cout<<"!!! "<<index<<" "<<dist[index]<<endl; edge.push_back(dist[index]); vis[index]=1; for(int j=2;j<=n;j++){ if( j!=index && dist[j]>a[index][j] ) dist[j] = a[index][j]; } } // for(int i=0;i<edge.size();i++) cout<<edge[i]<<" "; cout<<endl; sort(edge.begin(),edge.end()); reverse(edge.begin(),edge.end()); // for(int i=0;i<edge.size();i++) cout<<edge[i]<<" "; cout<<endl; printf("%.2lf\n",sqrt( double(edge[ s-1 ]) ) ); } return 0; }
POJ 3026 Borg Maze
bfs + kruskal
主要是建模
//#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<algorithm> #include<queue> #include<cstring> #include<map> #include<vector> #define inf 2e9 #define maxnode 200000 #define ll long long #define lowbit(x) (x&(-x)) const int mod = 1e9 + 7; const int maxn = 5e2 + 10; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; using namespace std; int par[maxn],cnt,mp[maxn][maxn]; struct edge{ int u,v,cost; edge(int u1=0,int v1=0,int c1=0): u(u1),v(v1),cost(c1) {} }edges[maxn*maxn]; bool cmp(edge n1,edge n2){ return n1.cost<n2.cost; } int find_root(int x){ if( x==par[x] ) return x; return par[x] = find_root( par[x] ); } int x[maxn],y[maxn],vis[maxn][maxn],n,m,mat[maxn][maxn]; char a[maxn][maxn]; struct node{ int x,y,step; node(int x1=0,int y1=0,int s1=0): x(x1),y(y1),step(s1) {} }; void bfs(int s){ queue<node> q; memset(vis,0,sizeof(vis)); q.push( node(x[s],y[s],0) ); vis[ x[s] ][ y[s] ]=1; while( !q.empty() ){ node u=q.front(); q.pop(); if( mp[u.x][u.y] ) mat[s][ mp[u.x][u.y] ] = u.step; for(int k=0;k<4;k++){ int x1=u.x+dx[k]; int y1=u.y+dy[k]; if( x1>=1 && x1<=n && y1>=1 && y1<=m && a[x1][y1]!=‘#‘ && vis[x1][y1]==0 ){ vis[x1][y1]=1; q.push( node(x1,y1,u.step+1) ); } } } } int main(){ int t,tc=0; scanf("%d",&t); while( t-- ){ scanf("%d%d",&m,&n); char temp[51]; gets(temp); memset(mp,0,sizeof(mp)); cnt=0; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ scanf("%c",&a[i][j]); if( a[i][j]==‘S‘ || a[i][j]==‘A‘ ) { x[++cnt]=i; y[cnt]=j; mp[i][j]=cnt; } } getchar(); } for(int i=1;i<=cnt;i++) bfs(i); for(int i=1;i<=cnt;i++) par[i]=i; int cnt_edge=0; for(int i=1;i<=cnt;i++) for(int j=i+1;j<=cnt;j++) edges[++cnt_edge] = edge(i,j, mat[i][j] ); sort(edges+1,edges+1+cnt_edge,cmp); int ans=0; for(int i=1;i<=cnt_edge;i++){ int u = edges[i].u,v=edges[i].v; int rootu = find_root(u), rootv = find_root(v); if( rootu!=rootv ){ par[rootu] = rootv; ans+=edges[i].cost; } } printf("%d\n",ans); } return 0; }
POJ 1679 The Unique MST
次小生成树
求出数组mat[u][v]表示最小生成树上u到v路径上的最长边边权，用dp的思路进行不断迭代，考验一定码力
//#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<algorithm> #include<cstring> #include<map> #include<cmath> #include<vector> #define inf 2e9 #define maxnode 200000 #define ll long long #define lowbit(x) (x&(-x)) const int mod = 1e9 + 7; const int maxn = 1e2 + 10; int dx[4]={0,0,1,-1}; int dy[4]={1,-1,0,0}; using namespace std; int mp[maxn][maxn],dist[maxn],pre[maxn]; int mat[maxn][maxn],used[maxn][maxn],vis[maxn];//mat[u][v] u到v最短路径的最长距离 int main(){ int t; cin>>t; while( t-- ){ int n,m; cin>>n>>m; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) mp[i][j]=inf; memset(used,0,sizeof(used)); memset(mat,0,sizeof(mat)); memset(vis,0,sizeof(vis)); for(int i=1;i<=m;i++){ int u,v,d; cin>>u>>v>>d; mp[u][v]=mp[v][u]=d; } vis[1]=1; for(int i=2;i<=n;i++) { dist[i]=mp[1][i]; pre[i]=1; } int ans=0; for(int i=2;i<=n;i++){ int mind=inf,index; for(int j=2;j<=n;j++){ if( !vis[j] && dist[j]<mind ) { mind=dist[j]; index=j; } } //找到要加入vis的顶点index used[index][pre[index]] = used[pre[index]][index]= 1; vis[index]=1; ans+=dist[index]; for(int j=2;j<=n;j++){ if( vis[j] ) mat[j][index] = mat[index][j] = max( mat[j][pre[index]],dist[index] ); if( !vis[j] && mp[index][j]<dist[j] ){ dist[j]=mp[index][j]; pre[j]=index; } } } //mst长度，并且mat维护出来 int minlen=inf; for(int i=1;i<=n;i++){ for(int j=i+1;j<=n;j++){ if( used[i][j]==0 && mp[i][j]!=inf ){ minlen = min( minlen,ans+mp[i][j]-mat[i][j] ); } } } if( minlen==ans ) cout<<"Not Unique!"<<endl; else cout<<ans<<endl; } return 0; }
标签：长度 void print possible min poj getc while distance
原文地址：https://www.cnblogs.com/ZhenghangHu/p/10259099.html
