BZOJ3832 [Poi2014]Rally 【拓扑序 + 堆】
2018-05-27 22:14
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题目链接
题解
神思路orz,根本不会做
设\(f[i]\)为到\(i\)的最长路,\(g[i]\)为\(i\)出发的最长路,二者可以拓扑序后\(dp\)求得
那么一条边\((u,v)\)的对应的最长链就是\(f[u] + 1 + g[v]\)
我们人为加入源汇点\(S\),\(T\),\(S\)向每个点连边,每个点向\(T\)连边
我们考虑把整个图划分开
一开始所有点都在\(T\)这边,割边为所有\(S\)的边
然后我们按照拓扑序把点逐一加入\(S\)集合中
加入时,我们删去\(S\)集合连向该点的边,然后询问所有边的最大值,即为删去该点的最长链
加入后,我们加入该点连向\(T\)集合的边
由于是按照拓扑序,所以以上提到的所有边就是该点的所有入边/出边
然后所有边的最大值可以用堆或者线段树维护
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<queue> #include<cmath> #include<map> #define Redge(u) for (int k = h[u]; k; k = ed[k].nxt) #define REP(i,n) for (int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,int>(a,b) #define cls(s) memset(s,0,sizeof(s)) #define cp pair<int,int> #define LL long long int using namespace std; const int maxn = 500005,maxm = 2000005,INF = 1000000000; inline int read(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();} return out * flag; } struct Heap{ priority_queue<int> a,b; void ck(){while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();} int size(){return a.size() - b.size();} void ins(int x){ck(); a.push(x);} void del(int x){ck(); b.push(x);} int top(){ck(); return size() ? a.top() : 0;} }H; int n,m,f[maxn],g[maxn],s[maxn]; int q[maxn],head,tail; int h[maxn],de[maxn],ne; int hi[maxn],nei; struct EDGE{int to,nxt;}ed[maxm],e[maxm]; inline void build(int u,int v){ ed[++ne] = (EDGE){v,h[u]}; h[u] = ne; de[v]++; e[++nei] = (EDGE){u,hi[v]}; hi[v] = nei; } void topu(){ head = 0; tail = -1; int u; REP(i,n) if (!de[i]) q[++tail] = i; REP(i,n){ s[i] = u = q[head++]; Redge(u) if (!(--de[ed[k].to])) q[++tail] = ed[k].to; } } void init(){ REP(i,n){ int u = s[i]; Redge(u) f[ed[k].to] = max(f[ed[k].to],f[u] + 1); } for (int i = n; i; i--){ int u = s[i]; Redge(u) g[u] = max(g[u],g[ed[k].to] + 1); } } void solve(){ int ans = INF,ansu = 0,x; REP(i,n) H.ins(g[i]); REP(i,n){ int u = s[i]; H.del(g[u]); for (int k = hi[u]; k; k = e[k].nxt) H.del(f[e[k].to] + 1 + g[u]); x = H.top(); if (x < ans) ans = x,ansu = u; H.ins(f[u]); Redge(u) H.ins(f[u] + 1 + g[ed[k].to]); } printf("%d %d\n",ansu,ans); } int main(){ n = read(); m = read(); int a,b; REP(i,m){ a = read(); b = read(); build(a,b); } topu(); init(); //REP(i,n) printf("node%d f = %d g = %d\n",i,f[i],g[i]); solve(); return 0; }
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