UVA - 10305 Ordering Tasks
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题目链接:http://vjudge.net/problem/UVA-10305
题意:设有n个变量标号1到n,共有m个二元组(u,v)表示变量u < v,请根据这m个二元组列出n个变量可能的大小关系。(答案不唯一)
分析:典型的拓扑排序。百度上拓扑排序的定义为:对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序,是将G中所有顶点排成一个线性序列,使得图中任意一对顶点u和v,若边(u,v)∈E(G),则u在线性序列中出现在v之前。方法:(1)用队列 (2)DFS遍历完成拓扑排序
方法一:对列
#include<cstdio>#include<cstdlib>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<set>#include<map>#include<vector>#include<list>#include<deque>#include<queue>#include<cmath>using namespace std;const int maxn = 110;int n,m;int d[maxn][maxn];//int vis[maxn];int drgee[maxn];int ans[maxn];void topo(){ queue<int> q; while(!q.empty()) q.pop(); for(int i = 1; i <= n; i++) if(!drgee[i]) q.push(i),drgee[i] = -1;//先将数为0的点加入队列中,将度数变为-1,防止再次将其加入队列中 int cur = 0; while(!q.empty()) { int tmp = q.front(); q.pop();//对队列首部元素 ans[cur++] = tmp;//ans用处储存结果 //printf("%d***\n",tmp); for(int i = 1; i <= n; i++) { if(d[tmp][i]) drgee[i]--;//将所有与之相连的点的入度数都减一 if(!drgee[i]) q.push(i),drgee[i] = -1;//若入度数为0,加入队列中,将度数标记为-1 } }}void print_ans(){ for(int i = 0; ans[i]; i++) { if(i) printf(" "); printf("%d",ans[i]); } printf("\n");}int main(){ while(scanf("%d%d",&n,&m) == 2) { if(!m && !n) break; memset(d, 0, sizeof(d)); // memset(vis, 0, sizeof(vis)); memset(drgee, 0, sizeof(drgee)); memset(ans, 0, sizeof(ans)); int u,v; for(int i = 0; i < m; i++) { scanf("%d%d",&u,&v); drgee[v]++;//记录每个点的入度数 d[u][v] = 1;//标记u v两点是否相连 } topo(); print_ans(); } return 0;}
方法二:DFS遍历
#include<cstdio>#include<cstdlib>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<set>#include<map>#include<vector>#include<list>#include<deque>#include<queue>#include<cmath>using namespace std;const int maxn = 110;int d[maxn][maxn];int vis[maxn];int ans[maxn];int n,m,t;bool dfs(int u){ vis[u] = -1;//标记u点正在被访问 for(int i = 1; i <= n; i++) { if(d[u][i]) { if(vis[i] == -1) return false;//说明存在环 if(!vis[i] && !dfs(i)) return false; } } vis[u] = 1;//标记u点已经被访问过了 ans[--t] = u;//将u点加入ans中,注意是倒着存储!!! return true;}bool topo(){ t = n; for(int i = 1; i <= n; i++) if(!vis[i])//为访问过i点 if(!dfs(i)) return false; return true;}void print_ans(){ for(int i = 0; ans[i]; i++) { if(i) printf(" "); printf("%d",ans[i]); } printf("\n");}int main(){ while(scanf("%d%d",&n,&m) == 2) { if(!m && !n) break; memset(d, 0, sizeof(d)); memset(vis, 0, sizeof(vis)); //memset(drgee, 0, sizeof(drgee)); memset(ans, 0, sizeof(ans));//ans用于储存结果 int u,v; for(int i = 0; i < m; i++) { scanf("%d%d",&u,&v); d[u][v] = 1;//标记u v相连 } topo(); print_ans(); } return 0;}
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