图的联通入门题
BZOJ 1051 [HAOI2006]受欢迎的牛
题意：每一头牛的愿望就是变成一头最受欢迎的牛。现在有N头牛，给你M对整数(A,B)，表示牛A认为牛B受欢迎。 这种关系是具有传递性的，如果A认为B受欢迎，B认为C受欢迎，那么牛A也认为牛C受欢迎。
你的任务是求出有多少头、牛被所有的牛认为是受欢迎的。
题解：强连通分量求缩点重构图（必定是DAG），出度为0的点为一个的时候输出该分量的大小，否则不存在。
#include <bits/stdc++.h> const int maxn=5e4+5; struct Edge{ int to, next; }edge[maxn], redge[maxn]; int head[maxn], tot; int rhead[maxn], cnt; int low[maxn], dfn[maxn]; int stk[maxn], instk[maxn], belong[maxn], num[maxn]; int idx, top, scc; int n, m; void addedge(int u, int v) { edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; } void tarjan(int u) { int v; low[u]=dfn[u]=++idx; stk[top++]=u, instk[u]=true; for(int i=head[u]; i!=-1; i=edge[i].next) { v=edge[i].to; if(!dfn[v]){ tarjan(v); low[u]=std::min(low[u], low[v]); } else if(instk[v]){ low[u]=std::min(low[u], dfn[v]); } } if(low[u]==dfn[u]) { scc++; do{ v=stk[--top]; instk[v]=false; belong[v]=scc; num[scc]++; }while(v!=u); } } void rebuild() { for(int u=1; u<=n; u++) for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to; if(belong[u]!=belong[v]) { redge[cnt].to=belong[v]; redge[cnt].next=rhead[belong[u]]; rhead[belong[u]]=cnt++; } } } int main() { scanf("%d%d", &n, &m); memset(head, -1, sizeof(head)); memset(rhead, -1, sizeof(rhead)); for(int i=0; i<m; i++) { int u, v; scanf("%d%d", &u, &v); addedge(u, v); } for(int u=1; u<=n; u++) if(!dfn[u]) tarjan(u); rebuild(); int ok=0, ans; for(int u=1; u<=scc; u++) if(rhead[u]==-1){ ok++; ans=num[u]; } printf("%d", ok==1? ans:0); return 0; }
仅记录缩点重构图的出度
1 #include <bits/stdc++.h> 2 3 const int maxn=5e4+5; 4 5 struct Edge{ 6 int to, next; 7 }edge[maxn], redge[maxn]; 8 int head[maxn], tot; 9 int out[maxn]; 10 11 int low[maxn], dfn[maxn]; 12 int stk[maxn], instk[maxn], belong[maxn], num[maxn]; 13 int idx, top, scc; 14 int n, m; 15 16 void addedge(int u, int v) 17 { 18 edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; 19 } 20 21 void tarjan(int u) 22 { 23 int v; 24 low[u]=dfn[u]=++idx; 25 stk[top++]=u, instk[u]=true; 26 for(int i=head[u]; i!=-1; i=edge[i].next) 27 { 28 v=edge[i].to; 29 if(!dfn[v]){ 30 tarjan(v); 31 low[u]=std::min(low[u], low[v]); 32 } 33 else if(instk[v]){ 34 low[u]=std::min(low[u], dfn[v]); 35 } 36 } 37 if(low[u]==dfn[u]) 38 { 39 scc++; 40 do{ 41 v=stk[--top]; 42 instk[v]=false; 43 belong[v]=scc; 44 num[scc]++; 45 }while(v!=u); 46 } 47 } 48 49 void rebuild() 50 { 51 for(int u=1; u<=n; u++) 52 for(int i=head[u]; i!=-1; i=edge[i].next) 53 { 54 int v=edge[i].to; 55 if(belong[u]!=belong[v]) 56 out[belong[u]]=1; 57 } 58 } 59 60 int main() 61 { 62 scanf("%d%d", &n, &m); 63 memset(head, -1, sizeof(head)); 64 for(int i=0; i<m; i++) 65 { 66 int u, v; scanf("%d%d", &u, &v); 67 addedge(u, v); 68 } 69 for(int u=1; u<=n; u++) 70 if(!dfn[u]) tarjan(u); 71 rebuild(); 72 int ok=0, ans; 73 for(int u=1; u<=scc; u++) 74 if(!out[u]){ 75 ok++; ans=num[u]; 76 } 77 printf("%d", ok==1? ans:0); 78 return 0; 79 }
BZOJ 1529 Ska Piggy Panks https://www.luogu.org/problem/P3420
题意：Byteazar 有 N 个小猪存钱罐. 每个存钱罐只能用钥匙打开或者砸开. Byteazar 已经把每个存钱罐的钥匙放到了某些存钱罐里，他想尽量少的打破存钱罐取出所有的钱,问最少要打破多少个存钱罐。给出：表示第i个存钱罐对应的钥匙放置在了第x个存钱罐中。
题解：1.tarjan缩点重构后求入度为0的点的个数，此题中会爆内存。
2.并查集直接搞就行。
#include <bits/stdc++.h> const int maxn=1e6+5; int fa[maxn]; int find(int x) { return fa[x]==x? x:fa[x]=find(fa[x]); } int main() { int n; scanf("%d", &n); for(int i=1; i<=n; i++) fa[i]=i; for(int i=1; i<=n; i++) { int x; scanf("%d", &x); int p=find(x), q=find(i); if(p!=q) fa[q]=p; } int ans=0; for(int i=1; i<=n; i++) if(fa[i]==i) ans++; printf("%d\n", ans); return 0; }
CF 427C Checkposts http://codeforces.com/problemset/problem/427/C
题意：条件：A checkpost at junction i can protect junction j if either i = j or the police patrol car can go to j from i and then come back to i. Building checkposts costs some money
求：You have to determine the minimum possible money needed to ensure the security of all the junctions. Also you have to find the number of ways to ensure the security in minimum price and in addition in minimum number of checkposts.
题解：tarjan的时候维护num【】，minw【】即可。ans1连加，ans2连乘
#include <bits/stdc++.h> const int maxn=3e5+5, mod=1e9+7; struct Edge{ int to, next; }edge[maxn]; int head[maxn], tot; int low[maxn], dfn[maxn]; int stk[maxn], instk[maxn], num[maxn]; int idx, top, scc; int n, m; int w[maxn], minw[maxn]; void addedge(int u, int v) { edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; } void tarjan(int u) { int v; low[u]=dfn[u]=++idx; stk[top++]=u, instk[u]=true; for(int i=head[u]; i!=-1; i=edge[i].next) { v=edge[i].to; if(!dfn[v]){ tarjan(v); low[u]=std::min(low[u], low[v]); } else if(instk[v]){ low[u]=std::min(low[u], dfn[v]); } } if(low[u]==dfn[u]) { scc++; do{ v=stk[--top]; instk[v]=false; if(minw[scc]>w[v]){ num[scc]=1; minw[scc]=w[v]; } else if(minw[scc]==w[v]) num[scc]++; }while(v!=u); } } int main() { memset(head, -1, sizeof(head)); memset(minw, 0x3f, sizeof(minw)); scanf("%d", &n); for(int i=1; i<=n; i++) scanf("%d", &w[i]); scanf("%d", &m); for(int i=0; i<m; i++) { int u, v; scanf("%d%d", &u, &v); addedge(u, v); } for(int u=1; u<=n; u++) if(!dfn[u]) tarjan(u); long long ans1=0, ans2=1; for(int i=1; i<=scc; i++) { ans1+=minw[i]; ans2=(ans2*num[i])%mod; } printf("%lld %lld\n", ans1, ans2); return 0; }
