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uva 10305 - Ordering Tasks
Problem F
Ordering Tasks
Input: standard input
Output: standard output
Time Limit: 1 second
Memory Limit: 32 MB
John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.
Input
The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and m. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.
Output
For each instance, print a line with n integers representing the tasks in a possible order of execution.
Sample Input
5 4
1 2
2 3
1 3
1 5
0 0
Sample Output
1 4 2 5 3
#include <iostream>#include <stack>#include <cstring>#include <cstdio>#include <string>#include <algorithm>#include <queue>#include <set>#include <map>#include <fstream>#include <stack>#include <list>#include <sstream>#include <cmath>using namespace std;#define ms(arr, val) memset(arr, val, sizeof(arr))#define mc(dest, src) memcpy(dest, src, sizeof(src))#define N 105#define INF 0x3fffffff#define vint vector<int>#define setint set<int>#define mint map<int, int>#define lint list<int>#define sch stack<char>#define qch queue<char>#define sint stack<int>#define qint queue<int>int ans[N], in[N], out[N], g[N][N], p;void init(){ p = 0; ms(in, 0); ms(out, 0); ms(g, 0);}qint qi;int main(){ int n, m, s, e; while (scanf("%d%d", &n, &m), n)//m可以为0 { init(); while (m--) { scanf("%d%d", &s, &e); in[e]++; out[s]++; g[s][e] = 1; } for (int i = 1; i <= n; i++) { if (!in[i] && !out[i]) { ans[p++] = i; } else if (!in[i]) { qi.push(i); } } while (!qi.empty()) { s = qi.front(); ans[p++] = s; qi.pop(); for (e = 1; e <= n; e++) { if (g[s][e]) { g[s][e] = 0; in[e]--; if (!in[e]) { qi.push(e); } } } } printf("%d", ans[0]); for (s = 1; s < p; s++) { printf(" %d", ans[s]); } putchar(‘\n‘); } return 0;}
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