hdu 2647 Reward - 拓扑排序

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

　　反向建图，跑拓扑排序即可。

Code

  1 /**
  2  * hdu
  3  * Problem#2647
  4  * Accepted
  5  * Time:31ms
  6  * Memory:1976k
  7  */
  8 #include <iostream>
  9 #include <cstdio>
 10 #include <ctime>
 11 #include <cmath>
 12 #include <cctype>
 13 #include <cstring>
 14 #include <cstdlib>
 15 #include <fstream>
 16 #include <sstream>
 17 #include <algorithm>
 18 #include <map>
 19 #include <set>
 20 #include <stack>
 21 #include <queue>
 22 #include <vector>
 23 #include <stack>
 24 #ifndef WIN32
 25 #define Auto "%lld"
 26 #else
 27 #define Auto "%I64d"
 28 #endif
 29 using namespace std;
 30 typedef bool boolean;
 31 const signed int inf = (signed)((1u << 31) - 1);
 32 const double eps = 1e-6;
 33 const int binary_limit = 128;
 34 #define smin(a, b) a = min(a, b)
 35 #define smax(a, b) a = max(a, b)
 36 #define max3(a, b, c) max(a, max(b, c))
 37 #define min3(a, b, c) min(a, min(b, c))
 38 template<typename T>
 39 inline boolean readInteger(T& u){
 40     char x;
 41     int aFlag = 1;
 42     while(!isdigit((x = getchar())) && x != '-' && x != -1);
 43     if(x == -1) {
 44         ungetc(x, stdin);    
 45         return false;
 46     }
 47     if(x == '-'){
 48         x = getchar();
 49         aFlag = -1;
 50     }
 51     for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
 52     ungetc(x, stdin);
 53     u *= aFlag;
 54     return true;
 55 }
 56 
 57 ///map template starts
 58 typedef class Edge{
 59     public:
 60         int end;
 61         int next;
 62         Edge(const int end = 0, const int next = 0):end(end), next(next){}
 63 }Edge;
 64 
 65 typedef class MapManager{
 66     public:
 67         int ce;
 68         int *h;
 69         Edge *edge;
 70         MapManager(){}
 71         MapManager(int points, int limit):ce(0){
 72             h = new int[(const int)(points + 1)];
 73             edge = new Edge[(const int)(limit + 1)];
 74             memset(h, 0, sizeof(int) * (points + 1));
 75         }
 76         inline void addEdge(int from, int end){
 77             edge[++ce] = Edge(end, h[from]);
 78             h[from] = ce;
 79         }
 80         inline void addDoubleEdge(int from, int end){
 81             addEdge(from, end);
 82             addEdge(end, from);
 83         }
 84         Edge& operator [] (int pos) {
 85             return edge[pos];
 86         }
 87         inline void clear() {
 88             delete[] edge;
 89             delete[] h;
 90         }
 91 }MapManager;
 92 #define m_begin(g, i) (g).h[(i)]
 93 ///map template ends
 94 
 95 int n, m;
 96 MapManager g;
 97 int* dag;
 98 int* dep;
 99 
100 inline boolean init() {
101     if(!readInteger(n))    return false;
102     readInteger(m);
103     g = MapManager(n, m);
104     dag = new int[(n + 1)];
105     dep = new int[(n + 1)];
106     memset(dag, 0, sizeof(int) * (n + 1));
107     memset(dep, -1, sizeof(int) * (n + 1));
108     for(int i = 1, a, b; i <= m; i++) {
109         readInteger(a);
110         readInteger(b);
111         g.addEdge(b, a);
112         dag[a]++;
113     }
114     return true;
115 }
116 
117 queue<int> que;
118 inline void topu() {
119     for(int i = 1; i <= n; i++)
120         if(!dag[i])
121             dep[i] = 888, que.push(i);
122     while(!que.empty()) {
123         int e = que.front();
124         que.pop();
125         for(int i = m_begin(g, e); i; i = g[i].next) {
126             int& eu = g[i].end;
127             dag[eu]--;
128             smax(dep[eu], dep[e] + 1);
129             if(!dag[eu])
130                 que.push(eu);
131         }
132     }
133 }
134 
135 int res;
136 inline void solve() {
137     res = 0;
138     topu();
139     for(int i = 1; i <= n; i++) {
140         if(dag[i]) {
141             puts("-1");
142             return;
143         }
144         res += dep[i];
145     }
146     printf("%d\n", res);
147 }
148 
149 inline void clear() {
150     g.clear();
151     delete[] dep;
152     delete[] dag;
153 }
154 
155 int main() {
156     while(init()) {
157         solve();
158         clear();
159     }
160     return 0;
161 }
posted @ 2017-07-10 12:00  阿波罗2003  阅读(167)  评论(0)  编辑  收藏  举报