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1258 Agri-Net (2)
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1164 The Castle (3)
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2395 Out of Hay (4)
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2337 Catenyms (7)
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2186 Popular Cows (8, challenge problem)
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1944 Fiber Communications (8, challenge problem)
poj2395
最小生成树的最大边
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; int fa[2010]; struct EDGE { int u, v, c; bool operator<(const EDGE &t)const { return c < t.c; } }; int find(int x) { if (x != fa[x]) fa[x] = find(fa[x]); return fa[x]; } EDGE edge[10010]; int main() { int n, m, a, b, c, i; scanf("%d%d", &n, &m); for (i = 1; i <= n; ++i) fa[i] = i; int cnt = 0; for (i = 0; i < m; ++i) { scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].c); } sort(edge, edge + m); for (i = 0; i < m; ++i) { a = find(edge[i].u); b = find(edge[i].v); if (a != b) { ++cnt; if (cnt == n - 1) break; fa[a] = b; } } printf("%d\n", edge[i].c); return 0; }
poj3177求双连通分量后缩点，判叶子节点数加一除二。
#include<cstdio> #include<cstring> #include<vector> #include<iostream> #include<algorithm> using namespace std; int d[5010],low[5010]; bool vis[5010][5010],mat[5010]; int m,n,tot; vector <int> adj[5010]; void dfs(int u,int p){ int v; mat[u]=1; vector <int>::iterator it; low[u]=tot++; for(it=adj[u].begin();it!=adj[u].end();++it){ v=*it; if(v==p) continue; if(!mat[v]) dfs(v,u); low[u]=min(low[u],low[v]); } } int main(){ // freopen("test.in","r",stdin); int i,u,v,ans; vector <int>::iterator it; while(scanf("%d%d",&n,&m)!=EOF){ memset(d,0,sizeof(d)); memset(vis,0,sizeof(vis)); for(i=1;i<=n;++i) adj[i].clear(); while(m--){ scanf("%d%d",&u,&v); if(vis[u][v] || vis[v][u]) continue; vis[u][v]=1; adj[u].push_back(v); adj[v].push_back(u); } tot=0; dfs(1,0); for(u=1;u<=n;++u){ for(it=adj[u].begin();it!=adj[u].end();++it){ v=*it; if(low[u]!=low[v]){ d[low[u]]++; } } } ans=0; for(u=0;u<n;++u){ if(d[u]==1) ans++; } printf("%d\n",(ans+1)>>1); } return 0; }
poj1945A*启发+广搜+优先队列
#include<cstdio> #include<iostream> #include<queue> #include<algorithm> #include<vector> using namespace std; #define maxn 80010 int n; void swap(int &a,int &b){ int t=a; a=b; b=t; } typedef struct S{ int x,y,s,c; S(){} S(int xx,int yy,int ss):x(xx),y(yy),s(ss){} S(const S& temp){ x=temp.x,y=temp.y,s=temp.s,c=temp.c; } bool operator<(const S t)const{ return c>t.c; } void out(){ printf("%d %d %d %d\n",x,y,s,c); } }NODE; NODE now,next; priority_queue <NODE> pq; typedef struct SS{ int x,y,s; SS(){} SS(int xx,int yy,int ss):x(xx),y(yy),s(ss){} }HASH; vector<HASH> hash[maxn]; void cal(NODE &node){ int temp=n/node.x,cnt=0; while(temp){ ++cnt; temp>>=1; } node.c=node.s+cnt; } bool gaoh(NODE &node){ int temp; temp=node.x+node.y; // printf("hash1111=");node.out(); vector<HASH>::iterator it; for(it=hash[temp].begin();it!=hash[temp].end();++it){ if(it->x==node.x && it->y==node.y){ if(it->s<=node.s) return 1; else { it->s=node.s; return 0; } } } hash[temp].push_back(HASH(node.x,node.y,node.s)); // printf("hash=");node.out(); return 0; } int gcd(int a,int b){ for(int t;t=b;b=a%b,a=t); return a; } bool gao(NODE &node){ // node.out(); if(node.x==n || node.y==n){ printf("%d\n",node.s); return 1; } if(node.x<node.y) swap(node.x,node.y); // printf("temp: "); node.out(); if(node.x<=0 || node.y<0 || (node.y==0 && node.x>n) || node.x==node.y || (node.x>n && node.y>n) || node.x>=2*n || n%gcd(node.x,node.y) || gaoh(node)) return 0; cal(node); pq.push(node); return 0; } int main(){ // freopen("aa.in","r",stdin); // freopen("aa.out","w",stdout); int i,x,y,s; while(scanf("%d",&n)!=EOF){ while(!pq.empty()) pq.pop(); for(i=0;i<maxn;++i) hash[i].clear(); if(n==1){ printf("0\n"); continue; } next=NODE(1,0,0); gaoh(next); cal(next); pq.push(next); while(!pq.empty()){ now=pq.top(); pq.pop(); x=now.x,y=now.y,s=now.s+1; // now.out(); next=NODE(x+x,y,s);if(gao(next)) break; next=NODE(x+x,x,s);if(gao(next)) break; next=NODE(x+y,y,s);if(gao(next)) break; next=NODE(x,x+y,s);if(gao(next)) break; next=NODE(x,y+y,s);if(gao(next)) break; next=NODE(y+y,y,s);if(gao(next)) break; next=NODE(x-y,y,s);if(gao(next)) break; next=NODE(x,y-x,s);if(gao(next)) break; next=NODE(x,x-y,s);if(gao(next)) break; next=NODE(y,y-x,s);if(gao(next)) break; } } return 0; }
poj2337Fleury算法求欧拉路
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> using namespace std; #define maxn 1010 bool h[30]; char sstr[maxn][22],*str[maxn]; int start[maxn],end[maxn]; int vis[maxn]; int l[30],r[30]; int rd[30],cd[30]; int n,des; struct cmp { bool operator () (const char *a, const char *b) { return strcmp(a, b)<0; } }; int fa[30]; int find(int x){ if(x!=fa[x]) fa[x]=find(fa[x]); return fa[x]; } int gao(int p,int have){ if(have==n){ des=p; return 1; } int i,next=end[p]; for(i=l[next];i<r[next];++i){ if(vis[i]==-1){ vis[i]=p; if(gao(i,have+1)) return 1; vis[i]=-1; } } return 0; } int ans[maxn]; int main(){ // freopen("test.in","r",stdin); int T,i,no,cnt,se; scanf("%d",&T); while(T--){ scanf("%d",&n); for(i=0;i<26;++i) l[i]=r[i]=rd[i]=cd[i]=0,fa[i]=i; for(i=0;i<n;++i){ scanf("%s",sstr[i]); str[i]=sstr[i]; } sort(str,str+n,cmp()); memset(h,0,sizeof(h)); for(i=0;i<n;++i){ start[i]=str[i][0]-'a'; end[i]=str[i][strlen(str[i])-1]-'a'; ++cd[start[i]]; ++rd[end[i]]; fa[find(start[i])]=find(end[i]); h[start[i]]=h[end[i]]=1; if(start[i]!=start[i-1]){ r[start[i-1]]=i; l[start[i]]=i; } } r[start[i-1]]=i; // for(i=0;i<n;++i) printf("%s\n",str[i]); int tot=0; for(i=0;i<26;++i) if(h[i] && find(i)==i) ++tot; if(tot>1){ printf("***\n"); continue; } cnt=0,no=0; for(i=0;i<26;++i){ if(rd[i]-cd[i]>1 || cd[i]-rd[i]>1){ no=1; break; } if(rd[i]-cd[i]==1){ ++cnt; } else if(cd[i]-rd[i]==1){ ++cnt; se=i; } } // printf("no=%d cnt=%d\n",no,cnt); if(no || cnt>2) { printf("***\n"); continue; } memset(vis,0xff,sizeof(vis)); if(!cnt){ vis[0]=-2; gao(0,1); } else{ for(i=l[se];i<r[se];++i){ vis[i]=-2; if(gao(i,1)) break; vis[i]=-2; } } cnt=0; int Des=des; do{ ans[cnt++]=des; des=vis[des]; } while(des!=-2); // for(i=0;i<cnt;++i) printf("ansi=%d\n",ans[i]); for(i=cnt-1;i>0;--i) printf("%s.",str[ans[i]]); printf("%s\n",str[ans[i]]); } return 0; }
poj2186求强连通分量后缩点，判唯一的出度为0的点的原始点数
#include<cstdio> #include<cstring> #include<stack> #include<vector> #include<iostream> #include<algorithm> using namespace std; #define maxn 10010 vector<int> adj[maxn]; int dfn[maxn],low[maxn],d[maxn],scc[maxn]; int tot,n,m,cnt; stack<int> s; void gao(int u){ int v; dfn[u]=low[u]=++tot; s.push(u); vector<int>::iterator it; for(it=adj[u].begin();it!=adj[u].end();++it){ v=*it; if(!dfn[v]) gao(v); low[u]=min(low[u],low[v]); } if(low[u]==dfn[u]){ ++cnt; while(!s.empty()){ v=s.top(); scc[v]=cnt; s.pop(); if(low[v]==dfn[v]) break; } } } int main(){ int u,v,i,z=0,ans,p; vector<int>::iterator it; scanf("%d%d",&n,&m); while(m--){ scanf("%d%d",&u,&v); adj[u].push_back(v); } for(i=1;i<=n;++i) if(!dfn[i]) gao(i); for(i=1;i<=n;++i){ for(it=adj[i].begin();it!=adj[i].end();++it){ if(scc[i]!=scc[*it]) ++d[scc[i]]; } } for(i=1;i<=cnt;++i) if(!d[i]) ++z,p=i; if(z>1 || !z) printf("0\n"); else{ ans=0; for(i=1;i<=n;++i) if(scc[i]==p) ++ans; printf("%d\n",ans); } return 0; }
poj1944枚举弃边，跳表统计数量
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define maxn 1010 #define maxp 10010 int n,p; int x[maxp],y[maxp]; int f[maxn]; void swap(int &a,int &b){ int t=a; a=b; b=t; } int main(){ int i,ans,j,temp,far; scanf("%d%d",&n,&p); ans=n; for(i=0;i<p;++i){ scanf("%d%d",&x[i],&y[i]); if(x[i]==y[i]){ --i,--p; }else if(x[i]>y[i]){ swap(x[i],y[i]); } } for(i=1;i<=n;++i){ memset(f,0,sizeof(f)); for(j=0;j<p;++j){ if(x[j]>=i+1 || y[j]<=i){ f[x[j]]=max(y[j],f[x[j]]); }else{ f[y[j]]=n+1; f[1]=max(x[j],f[1]); } } temp=0; far=1; for(j=1;j<=n;++j){ if(f[j]>far){ temp+=f[j]-max(far,j); far=f[j]; } } ans=min(temp,ans); } printf("%d\n",ans); return 0; }
Assignment 6: Basic Graph Algorithms
最新推荐文章于 2025-10-18 16:05:59 发布
本文精选了五道ACM竞赛中的经典算法题目，并详细解析了每道题目的解题思路与实现代码，包括最小生成树的最大边、求双连通分量、A*启发式搜索、Fleury算法求欧拉路径及强连通分量的应用。
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