[POJ3249]Test for Job [拓扑排序+DAG上的最长路径]

给定一张带点权的DAG 求一条入度为0节点到出度为0节点的最长路

把点权转化为边权（同时多源转化成单源）：边u->v的权值为W[v]，这样入度为0的节点权值会被遗漏，新开一个点0向入度为0的点u连有向边，权值为W[u]，这样就只有0是入度为0的点了。

先进行拓扑排序，再求出最长路径（利用DAG分层的性质可以在线性时间内求出最长路）

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long LL;

template<class T> void read(T & x){
  register int c = getchar(), f = 1; x = 0;
  while(!isdigit(c)){if (c == '-') f = -1; c = getchar();}
  while(isdigit(c)) x = x*10+(c&15), c = getchar();
  x *= f;
}
const int N = 100005;
struct Edge{int v, w, next;}G[N*20];
int n, m, s, head[N], tot,  ind[N], out[N], W[N]; LL dis[N], ans = -123023423423;
void toposort(int s) {
  int topo[N], cnt = 0;
  topo[++cnt] = 0;
  for(int k = 0; k <= cnt; ++k)
    for(int i = head[topo[k]]; i; i = G[i].next) {
      int v = G[i].v;
      ind[v]--;
      if (ind[v]==0) topo[++cnt] = v;
    }
  memset(dis,0xc0,sizeof(dis));dis[s] = 0;
  for(int k = 0; k <= cnt; ++k)
    for(int i = head[topo[k]]; i; i = G[i].next) {
      int v = G[i].v, w = G[i].w;
      dis[v] = max(dis[topo[k]]+w, dis[v]);
    }
}

inline void add(int u, int v, int w) {
  G[++tot].v=v, G[tot].w=w, G[tot].next=head[u];head[u]=tot;
  ind[v]++, out[u]++;
}

int main(void){
  while(scanf("%d%d", &n, &m) == 2){
    memset(head,0,sizeof(head));tot=0;memset(ind,0,sizeof(ind));memset(out,0,sizeof(out));
    ans = 0xc0c0c0c0c0c0c0c0;
    for(int i = 1; i <= n; ++i) read(W[i]);
    for(int u, v, i = 1; i <= m; ++i) {
      read(u), read(v);
      add(u, v, W[v]);
    } 
    for(int i = 1; i <= n; ++i) {
      if (!ind[i]) add(0,i,W[i]);
    }
    toposort(s);
    for(int i = 1; i <= n; ++i) {
      if (!out[i]) ans = max(ans, dis[i]);
    }
    cout << ans << '
';
  }
  return 0;
}