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

输入样例#1：

复制

2 2
1 1
1 2
2 1

输出样例#1： 复制

2

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
}

 

EPFL - Fighting