题目背景
缩点+DP
题目描述
给定一个n个点m条边有向图，每个点有一个权值，求一条路径，使路径经过的点权值之和最大。你只需要求出这个权值和。
允许多次经过一条边或者一个点，但是，重复经过的点，权值只计算一次。
输入输出格式
输入格式：第一行，n,m
第二行，n个整数，依次代表点权
第三至m+2行，每行两个整数u,v，表示u->v有一条有向边
输出格式：共一行，最大的点权之和。
输入输出样例
说明
n<=10^4,m<=10^5,点权<=1000
算法：Tarjan缩点+DAGdp
Tarjan+记忆化搜索；
缩点以后，重新建图，然后dp；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 400005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n, m;
int idx;
int col[maxn], dp[maxn], sum[maxn];
int head[maxn];
int sk[maxn], top;
int dfn[maxn], low[maxn];
int tot;
int vis[maxn];
int val[maxn];
struct node {
int u, v, nxt;
}edge[maxn];
int cnt;
void addedge(int x, int y) {
edge[++cnt].v = y; edge[cnt].nxt = head[x]; head[x] = cnt;
}
void tarjan(int x) {
sk[++top] = x; vis[x] = 1;
low[x] = dfn[x] = ++idx;
for (int i = head[x]; i; i = edge[i].nxt) {
int v = edge[i].v;
if (!dfn[v]) {
tarjan(v);
low[x] = min(low[x], low[v]);
}
else if (vis[v]) {
low[x] = min(low[x], dfn[v]);
}
}
if (dfn[x] == low[x]) {
tot++;
while (sk[top + 1] != x) {
col[sk[top]] = tot; sum[tot] += val[sk[top]]; vis[sk[top--]] = 0;
}
}
}
void DP(int x) {
int maxx = 0;
if (dp[x])return;
dp[x] = sum[x];
for (int i = head[x]; i; i = edge[i].nxt) {
int v = edge[i].v;
if (!dp[v])DP(v);
maxx = max(maxx, dp[v]);
}
dp[x] += maxx;
}
int x[maxn], y[maxn];
int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m);
for (int i = 1; i <= n; i++)rdint(val[i]);
for (int i = 1; i <= m; i++) {
rdint(x[i]); rdint(y[i]); addedge(x[i], y[i]);
}
for (int i = 1; i <= n; i++)if (!dfn[i])tarjan(i);
ms(edge); cnt = 0; ms(head);
for (int i = 1; i <= m; i++) {
if (col[x[i]] != col[y[i]]) {
addedge(col[x[i]], col[y[i]]);
}
}
int ans = 0;
for (int i = 1; i <= tot; i++) {
if (!dp[i]) {
DP(i); ans = max(ans, dp[i]);
}
}
cout << ans << endl;
return 0;
}