题意 给出一些任务的优先级别 将这些任务进行的时间 进行先后排序
思路 拓扑排序
将所以有先后关系的任务都连一条边
然后每次 输出 度为0 的任务 每次把 以这个任务为弧的边 都取消 相对应任务的度也-1
再循环
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = acos(-1.0); const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 1e2 + 5; const int MOD = 1e9 + 7; int G[maxn][maxn]; int degree[maxn]; int v[maxn]; int n, m; vector <int> ans; int Count; void dfs() { for (int i = 1; i <= n; i++) { if (degree[i] == 0 && v[i] == 0) { Count++; ans.pb(i); v[i] = 1; for (int j = 1; j <= n; j++) { if (G[i][j]) { G[i][j] = 1; degree[j]--; } } } } if (Count != n) dfs(); } int main() { while (scanf("%d%d", &n, &m) && (n || m)) { CLR(G); CLR(degree); CLR(v); int x, y; for (int i = 0; i < m; i++) { scanf("%d%d", &x, &y); G[x][y] = G[y][x] = 1; degree[y]++; } ans.clear(); Count = 0; dfs(); vector <int>::iterator it; for (it = ans.begin(); it != ans.end(); it++) { if (it != ans.begin()) printf(" "); printf("%d", (*it)); } cout << endl; } }转载于:https://www.cnblogs.com/Dup4/p/9433130.html
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