P1113 杂务

输入输出样例

输入 #1

7
1 5 0
2 2 1 0
3 3 2 0
4 6 1 0
5 1 2 4 0
6 8 2 4 0
7 4 3 5 6 0

输出 #1

23

思路
　　由主次关系可想到拓扑排序，跑一遍拓扑排序得到一种线性的工作方式，维护每个任务点时间花费的最大值即可

CODE

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int maxn = 1e5 + 7;

int in[maxn], out[maxn];
int val[maxn];
int cost[maxn];
int edge[maxn], head[maxn], nxt[maxn], cnt;
int n, ans;

void BuildGraph(int u, int v) {
    cnt++;
    edge[cnt] = v;
    nxt[cnt] = head[u];
    head[u] = cnt;
}

void topo() {
    queue <int> q;
    for ( int i = 1; i <= n; ++i ) {
        if(in[i] == 0) {
            q.push(i);
            cost[i] = val[i];
        }
    }
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i; i = nxt[i]) {
            int v = edge[i];
            cost[v] = max(cost[v], cost[u] + val[v]);
            --in[v];
            if(!in[v]) {
                if(!out[v]) {
                    ans = max(ans, cost[v]);
                }
                else {
                    q.push(v);
                }
            }
        }
    }
    
}
int main()
{
    read(n);
    for ( int i = 1; i <= n; ++i ) {
        int id = 0, c = 0;
        scanf("%d %d",&id, &c);
        val[id] = c;
        int x;
        while(scanf("%d",&x) && x) {
            BuildGraph(id, x);
            out[id]++;
            in[x]++;
        }
    }
    topo();
    printf("%d
",ans);
    return 0;
}