- 题目大意
给你一个图，求一条起点（入度为0）到终点（出度为0）的路。满足全部点的val之和最大。
- 解题思路
先用数组记录入度为0和出度为0的点，然后利用记忆化搜索来解决。
- 代码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f;
const int N = 1e6 + 10;
long long cnt,sum;
long long d[N], head[N], num[N], vis[N];
struct egde {
long long u, v, w;
}e[N];
long long dfs(long long u)
{
if (d[u])
return d[u];
long long MAX = -INF;
for (long long i = head[u]; ~i; i = e[i].w)
{
long long v = e[i].v;
MAX = max(MAX, dfs(v));
}
if (MAX== -INF)
MAX = 0;
d[u] =MAX+ num[u];
return d[u];
}
void addEdge(long long u, long long v)
{
e[cnt].v = v;
e[cnt].u = u;
e[cnt].w = head[u];
head[u] = cnt++;
}
int main()
{
long long n, m, a, b;
while (scanf_s("%lld%lld", &n, &m) != EOF)
{
for (long long i = 1; i <= n; i++)
scanf_s("%lld", &num[i]);
cnt = 0;
memset(head, -1, sizeof(head));
memset(vis, 0, sizeof(head));
memset(d, 0, sizeof(d));
for (long long i = 0; i < m; i++)
{
scanf_s("%lld%lld", &a, &b);
addEdge(a, b);
vis[b]++;
}
sum = -INF;
for (long long i = 1; i <= n; i++)
{
if (vis[i] == 0)
sum = max(sum, dfs(i));
}
printf("%lld\n", sum);
}
return 0;
}