HDU 2647 Reward（图论-拓扑排序）

Reward

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 11 22 21 22 1

 

Sample Output

1777-1

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

Recommend

yifenfei

题目大意：

n个人，m条边，每条边a，b 表示a比b的工资要多，每个人的工资至少888，问满足关系的工资总和至少多少？如果出现关系矛盾，输出-1

解题思路：

根据工资关系建立拓扑图，0入度的人工资从888开始，一层一层，逐渐增加工资，若最后还有人入度不为0，则出现矛盾。

参考代码：

#include <iostream>#include <algorithm>#include <cstring>#include <queue>using namespace std;const int MAXN = 10010;int inDegree[MAXN], ans, cnt, n, m;vector<int> child[MAXN];struct State {    int reward, node;    State(int _reward, int _node) : reward(_reward), node(_node) {}};queue<State> q;void init() {    memset(inDegree, 0, sizeof(inDegree));    for (int i = 0; i <= n; i++) {        child[i].clear();    }    ans = cnt = 0;    while (!q.empty()) q.pop();}void input() {    for (int i = 0; i < m; i++) {        int a, b;        cin >> a >> b;        inDegree[a]++;        child[b].push_back(a);    }}void work() {    for (int i = 1; i <= n; i++) {        if (inDegree[i] == 0) {            q.push(State(888, i));            cnt++;        }    }    while (!q.empty()) {        State cur = q.front();        q.pop();        ans += cur.reward;        for (int i = 0; i < child[cur.node].size(); i++) {            inDegree[child[cur.node][i]]--;            if (inDegree[child[cur.node][i]] == 0) {                q.push(State(cur.reward+1, child[cur.node][i]));                cnt++;            }        }    }}void output() {    if (cnt == n) {        cout << ans << endl;    } else {        cout << -1 << endl;    }}int main() {    ios::sync_with_stdio(false);    while (cin >> n >> m) {        init();        input();        work();        output();    }    return 0;}