HDU Problem 2647 Reward【拓扑排序】

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7391    Accepted Submission(s): 2317

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

 

思路：将图倒着存，如果找到一个节点的上一级，就用res数组保存这个节点上级所处的等级，最后做一个统计。

#include <bits/stdc++.h>#define MAXN 10005using namespace std;int n, m, outdegree[MAXN], res[MAXN];int head[MAXN], num, que[MAXN];struct node{    int from, to, next;}edge[MAXN*2];void add(int x, int y) {    edge[num].from = x;    edge[num].to = y;    edge[num].next = head[x];    head[x] = num++;}void init() {    memset(head, -1, sizeof(head));    memset(outdegree, 0, sizeof(outdegree));    memset(res, 0, sizeof(res));    num = 0;}void solved() {    int iq = 0, ans = 0;    for (int i = 1; i <= n; i++) {        if (outdegree[i] == 0) {            que[iq++] = i;        }    }    for (int i = 0; i < iq; i++) {        for (int k = head[que[i]]; k != -1; k = edge[k].next) {            outdegree[edge[k].to]--;            if (outdegree[edge[k].to] == 0) {                que[iq++] = edge[k].to;                //记录该节点所在的等级                res[edge[k].to] = res[edge[k].from] + 1;            }        }    }    for (int i = 1; i <= n; i++) {        ans += res[i];        //如果存在环        if (outdegree[i] > 0) {            printf("-1\n"); return ;        }    }    printf("%d\n", ans + 888*n);}int main() {    int a, b;    while (scanf("%d%d", &n, &m) != EOF) {        init();        for (int i = 0; i < m; i++) {            scanf("%d%d", &a, &b);            add(b, a); outdegree[a]++;        }        solved();    }    return 0;}