拓扑排序 -- Codeforces Round #660 (Div. 2) D Captain Flint and Treasure

D Captain Flint and Treasure

题意：

给出两个序列a和b，一次操作你可以将ans+=a[i],a[b[i]]+=a[i]。询问最后ans的最大值。

题解：

拓扑排序，遇到负权值的节点就跳过，减少它给答案带来的贡献。

code：

#include
using namespace std;
const int maxn = 2e5 + 5;
long long a[maxn], ans;
int b[maxn], c[maxn], d[maxn];
bool e[maxn];
vector<int> v;
void init(int n){
	v.clear();
	for(int i = 1; i <= n; i++) {
		c[i] = 0;
		e[i] = false;
	}
	
} 

int main(){
	int n;
	scanf("%d", &n);
	
	init(n);
	
	int x;
	for(int i = 1; i <= n; i++){
		scanf("%d", &x);
		a[i] = (long long)x;
	}
	for(int i = 1; i <= n; i++) {
		scanf("%d", &b[i]);
		if(b[i] != -1) c[b[i]]++;
	}
	
	int k = 0;
	for(int i = 1; i <= n; i++) {
		if(c[i] == 0) d[++k] = i; 
	} 
	
	ans = 0;
	for(int i = 1; i <= n; i++){
		ans += a[d[i]];
		if(b[d[i]] != -1){
			if(a[d[i]] > 0){
				a[b[d[i]]] += a[d[i]];
				v.push_back(d[i]);
			    e[d[i]] = true;			
			}
			
		    c[b[d[i]]]--;
		    
		    if(c[b[d[i]]] == 0) d[++k] = b[d[i]];
		} 	
				
	}
	for(int i = n; i >= 1; i--) if(!e[d[i]]) v.push_back(d[i]);
	printf("%lld\n", ans);
	for(int i = 0; i < v.size(); i++) printf("%d ", v[i]);
	printf("\n"); 
	
}

dalao的代码：

#include
using namespace std;
const int maxn=2e5+100;
typedef long long ll;
ll a[maxn];
int b[maxn];
int inDegree[maxn];
int n;
vector<int> g[maxn];
ll ans;
void topo () {
    queue<int> q;
    for (int i=1;i<=n;i++) 
        if (inDegree[i]==0) {
            q.push(i);
        }
    
    while (!q.empty()) {
        int u=q.front();
        q.pop();
        ans+=a[u];
        if (b[u]==-1) continue;
        if (a[u]>0) {
            g[u].push_back(b[u]);
            a[b[u]]+=a[u];
        }
        else {
            g[b[u]].push_back(u);
        }
        if (--inDegree[b[u]]==0) q.push(b[u]);
    }
    
}
int main () {
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lld",a+i);
    for (int i=1;i<=n;i++) {
        scanf("%d",b+i);
        if (b[i]>0) {
            inDegree[b[i]]++;
        }
    }
    topo();
    printf("%lld\n",ans);
    vector<int> wjm;
    for (int i=1;i<=n;i++)
        for (int v:g[i]) inDegree[v]++;
    for (int i=1;i<=n;i++) if (inDegree[i]==0) wjm.push_back(i);
    for (int i=0;i<n;i++) {
        printf("%d ",wjm[i]);
        for (int v:g[wjm[i]]) 
            if (--inDegree[v]==0) wjm.push_back(v);
    }
}