標籤:eps stack lib ack map const iostream task while
題意:給定優先關係進行拓撲排序。
分析:將入度為0的點加入優先隊列,並將與之相連的點入度減1,若又有度數為0的點,繼續加入優先隊列,依次類推。
#pragma comment(linker, "/STACK:102400000, 102400000")#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define Min(a, b) ((a < b) ? a : b)#define Max(a, b) ((a < b) ? b : a)typedef long long ll;typedef unsigned long long llu;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const ll LL_INF = 0x3f3f3f3f3f3f3f3f;const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const double eps = 1e-8;const int MAXN = 100 + 10;const int MAXT = 10000 + 10;using namespace std;int in[MAXN];vector<int> a[MAXN];vector<int> ans;priority_queue<int, vector<int>, greater<int> > q;int main(){ int n, m; while(scanf("%d%d", &n, &m) == 2){ if(!n && !m) return 0; memset(in, 0, sizeof in); ans.clear(); for(int i = 0; i < MAXN; ++i) a[i].clear(); while(m--){ int x, y; scanf("%d%d", &x, &y); a[x].push_back(y); ++in[y]; } for(int i = 1; i <= n; ++i){ if(in[i] == 0){ q.push(i); } } while(!q.empty()){ int t = q.top(); q.pop(); ans.push_back(t); int len = a[t].size(); for(int i = 0; i < len; ++i){ if(--in[a[t][i]] == 0){ q.push(a[t][i]); } } } int len = ans.size(); for(int i = 0; i < len; ++i){ if(i) printf(" "); printf("%d", ans[i]); } printf("\n"); } return 0;}
UVA - 10305 Ordering Tasks(拓撲排序)