洛谷 - P3627 [APIO2009] -- 抢掠计划【缩点 + 最长路】

思路

1.先缩点，求出每个强连通分量的金钱数
2.然后建立缩点后的关系以及权值，跑一个模板最长路(路径取反，跑dijkstra(没有环)或者spfa)。

AC代码

/**
	缩点 + spfa走最长路 
*/
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef pair<int, int> pii;
#define IOS ios::sync_with_stdio(false)
const int maxn = 500050;
const int inf = 0x3f3f3f3f;
struct node{
	int to, w;
};
int dfn[maxn], low[maxn], color[maxn], sum[maxn], tot, ind = 1;
int from[maxn], to[maxn], cost[maxn], dis[maxn];
bool instack[maxn], vis[maxn];
stack<int> st;
vector<int> G[maxn]; // 记录tarjan
vector<node> G1[maxn]; // 记录spfa 
void tarjan(int u){
	dfn[u] = low[u] = ind++;
	st.push(u);
	instack[u] = true;
	for (int v : G[u]){
		if (!dfn[v]){
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if (instack[v]){
			low[u] = min(low[u], dfn[v]);
		}
	}
	if (low[u] == dfn[u]){
		int now;
		tot++;
		do{
			now = st.top(); st.pop();
			instack[now] = false;
			color[now] = tot;
			sum[tot] += cost[now];
		}while (now != u);
	}
}
void spfa(int st){
	queue<int> q;
	for (int i = 1; i <= tot; ++i) dis[i] = inf;
	st = color[st];
	q.push(st);
	dis[st] = -sum[st];
	while (!q.empty()){
		int u = q.front(); q.pop();
		vis[u] = false;
		for (auto i : G1[u]){
			int v = i.to;
			if (dis[v] > dis[u] + i.w){
				dis[v] = dis[u] + i.w;
				if (!vis[v]){
					vis[v] = true;
					q.push(v);
				}	
			}
		}
	}
}
void solve(){
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; ++i){
		scanf("%d%d", &from[i], &to[i]);
		G[from[i]].push_back(to[i]);
	}
	for (int i = 1; i <= n; ++i)	scanf("%d", &cost[i]);
	for (int i = 1; i <= n; ++i){
		if (!dfn[i]){
			tarjan(i);
		}
	}
	for (int i = 1; i <= m; ++i){
		if(color[from[i]] != color[to[i]]){
			G1[color[from[i]]].push_back(node{color[to[i]], -sum[color[to[i]]]});
		}
	}
	int p, x, st;
	scanf("%d%d", &st, &p);
	spfa(st);
	int ans = 0;
	while (p--){
		scanf("%d", &x);
		if (ans < -dis[color[x]]){
			ans = -dis[color[x]];
		}
	}
	printf("%d\n", ans);
}
int main(){
	solve();
	return 0;
}