HDU-2647 Reward（链式前向星+拓扑排序）

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16602 Accepted Submission(s): 5308

Problem Description

Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.

Input

One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.

Output

For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.

Sample Input

2 1
1 2
2 2
1 2
2 1

Sample Output

1777
-1

解题思路：

反向建图后拓扑排序，就为了复习下链式前向星和拓扑排序

#include <bits/stdc++.h>
using namespace std;
/*    freopen("k.in", "r", stdin);
    freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 5e5 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
int n, m;
int cnt = -1;
int indegree[MAXN];
int head[MAXN];
int red[MAXN];
struct Edge
{
    int u, v, Next;
    Edge(int uu = 0, int vv = 0, int NN = 0) { u = uu, v = vv, Next = NN; }
} e[MAXN];
void add(int u, int v)
{
    e[++cnt].v = v;
    e[cnt].u = u;
    e[cnt].Next = head[u];
    head[u] = cnt;
}
ll topo()
{
    int tot = 0;
    queue<int> q;
    bool flag1 = false;
    for (int i = 1; i <= n; i++)
        if (!indegree[i])
            q.push(i);
    while (!q.empty())
    {
        int now = q.front();
        q.pop();
        tot++;
        if (tot > n)
            break;
        for (int i = head[now]; ~i; i = e[i].Next)
        {
            int v = e[i].v;
            indegree[v]--;
            if (!indegree[v])
            {
                q.push(v);
                red[v] = red[now] + 1;
            }
        }
    }
    if (tot != n)
        return -1;
    ll ans = n * 888;
    for (int i = 1; i <= n; i++)
        ans += red[i];
    return ans;
}
int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        cnt = -1;
        memset(red, 0, sizeof(red));
        memset(head, -1, sizeof(head));
        memset(indegree, 0, sizeof(indegree));
        while (m--)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            add(v, u);
            indegree[u]++;
        }
        printf("%lld
", topo());
    }
    return 0;
}