uva10305 拓扑排序

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John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task isonly possible if other tasks have already been executed.

题干很短，意思很简单，拓扑排序。

我的方法是用队列做，将入度为零的点进队，每次出队更新入度，继续将入度为零的点进队。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <vector>#include <queue>using namespace std;typedef long long ll;vector<int> mp[105];vector<int> ans;int ru[105];int n,m;void init(){    for(int i=0; i<=n; ++i)    {        mp[i].clear();    }    memset(ru, 0, sizeof(ru));}int main(){int x,y;while(scanf("%d %d",&n,&m)&&(n||m))        {                init();                ans.clear();                queue<int> q;                for(int i=0;i<m;i++)                {                        scanf("%d %d",&x,&y);                        mp[x].push_back(y);                        ++ru[y];                }                for(int i=1; i<=n; i++)                if(!ru[i]) q.push(i);                while(!q.empty())                {                        int t=q.front();                        q.pop();                        ans.push_back(t);                        for(int i=0;i<mp[t].size();i++)                        {                                --ru[mp[t][i]];                                if(ru[mp[t][i]]==0)                                        q.push(mp[t][i]);                        }                }                printf("%d",ans[0]);                for(int i=1;i<ans.size();i++)                {                        printf(" %d",ans[i]);                }                printf("\n");        }}

貌似还有一种用dfs做的方法，思路也蛮清晰的

就不贴出来了

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