洛谷P3387缩点

传送门

有向图。。

代码中有两种方法，拓扑排序和记忆化搜索

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define re register
using namespace std ;
const int maxn = 1e5 + 4 ;

inline int read () {
	int f = 1 , x = 0 ;
	char ch = getchar () ;
	while(ch > '9' || ch < '0')  {if(ch == '-')  f = -1 ; ch = getchar () ;}
	while(ch >= '0' && ch <= '9')  {x = (x << 1) +(x << 3) + ch - '0' ; ch = getchar () ;}
	return x * f;
}

int n , m , a[maxn] , u , v , ans ;
int head[maxn] , tot , head2[maxn] , tot2 ;

struct Edge {
	int from , to , next ;
}edge[maxn << 1] , edge2[maxn] ;

inline void add(int u , int v) {
	edge[++tot].from = u ;
	edge[tot].to = v ;
	edge[tot].next =head[u] ;
	head[u] = tot ;
}

inline void add2(int u , int v) {
	edge2[++tot2].from = u ;
	edge2[tot2].to = v;
	edge2[tot2].next = head2[u];
	head2[u] = tot2;
}

int stack[maxn] , belong[maxn] , top , cnt , num[maxn] ;
int low[maxn] , dfn[maxn] , ind , dis[maxn] ;
bool ins[maxn] ;

inline void tarjan(int x) {
    dfn[x] = low[x] = ++ind;
    stack[++top] = x ;
    ins[x] = true ;
    for(re int i = head[x] ; i; i = edge[i].next) {
        int v = edge[i].to ;
        if(!dfn[v]) {
            tarjan(v) ;
            low[x] = min(low[x] , low[v]);
        }
        if(ins[v]) {
            low[x] = min(low[x] , dfn[v]);
        }
    }
    int k = 0;
    if(dfn[x] == low[x]) {
        ++cnt;
        do{
            k = stack[top];
            top--;
            ins[k] = false;
            num[cnt]++;
            dis[cnt] += a[k];
            belong[k] = cnt;
        }while (k != x);
    }
}

queue<int> q;
int dis2[maxn] , in[maxn] ;

inline int topo() {
    for(re int i = 1 ; i <= cnt ; ++ i)
    	if(!in[i]) {
    		q.push(i);
    		dis2[i] = dis[i];
    	}
    while(!q.empty()) {
    	int cur = q.front();
    	q.pop() ;
    	for(re int i = head2[cur] ; i ; i = edge2[i].next) {
    		int v = edge2[i].to ;
    		dis2[v] = max(dis2[v] , dis2[cur] + dis[v]) ;
    		in[edge2[i].to]-- ;
    		if(!in[edge2[i].to])
    			q.push(edge2[i].to) ;
    	}
    }
    for(re int i = 1 ; i <= n ; ++ i)
    	ans = max(ans , dis2[i]) ;
    return ans;
}

int main () {
	n = read () ; m =read() ;
	for(re int i = 1 ; i <= n ; ++ i)
		a[i] = read () ;
	for(re int i = 1 ; i <= m ; ++ i) {
		u = read () ; v = read () ;
		add(u , v) ;
	}
	for(re int i = 1 ; i <= n ; ++ i)
		if(!dfn[i])  tarjan(i) ;
	for(re int i = 1 ; i <= tot ; ++ i) {
        if(belong[edge[i].from] != belong[edge[i].to]) {
            add2(belong[edge[i].from] , belong[edge[i].to]) ;
            in[belong[edge[i].to]]++ ;
        }
    }
	printf("%d\n" , topo()) ;
	return 0 ;
}