拓扑排序

1.家谱树

寻寻有一个大家庭，辈分关系很混乱，请你帮他梳理一下家庭成员的关系。

输入：

第一行n，表示共5个人。

接下来的n行，第i行表示第i个人的孩子。

每行以0结束。

输出：

一行序列，空格隔开，使得每个人的后辈都比那个人后出现。

spj

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<stack>
using namespace std;
int n,cnt,ru[10005],hd[10005],ans, x,t,y;
stack<int> q;
struct Edge{
    int to, nxt;
}edge[1000005];
void add(int u, int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
int main(){
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        while(~scanf("%d",&y)){
            if(y == 0) break;
            add(i,y);
            ru[y]++;
        }
    }
    int mx = 0;
    for(int i = 1; i <= n; i++)
        if(ru[i] == 0){
             q.push(i);
        }
    while(!q.empty()){
        int u = q.top(); q.pop();
        printf("%d ",u);
        for(int i = hd[u]; i; i = edge[i].nxt){
            int v = edge[i].to;
            ru[v]--;
            if(!ru[v]){
                q.push(v);
            }
        }
    }
    return 0;
}
/*
5
0
4 5 0
1 0
5 3 0

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<stack>
using namespace std;
int n,cnt,ru[10005],hd[10005],ans, x,t,y;
stack<int> q;
struct Edge{
    int to, nxt;
}edge[1000005];
void add(int u, int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
int main(){
    freopen("t.in","r",stdin);
    freopen("t.out","w",stdout);
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        while(~scanf("%d",&y)){
            if(y == 0) break;
            add(i,y);
            ru[y]++;
        }
    }
    int mx = 0;
    for(int i = 1; i <= n; i++)
        if(ru[i] == 0){
             q.push(i);
        }
    while(!q.empty()){
        int u = q.top(); q.pop();
        printf("%d ",u);
        for(int i = hd[u]; i; i = edge[i].nxt){
            int v = edge[i].to;
            ru[v]--;
            if(!ru[v]){
                q.push(v);
            }
        }
    }
    return 0;
}
/*
5
0
4 5 0
1 0
5 3 0
3 0
*/

P1113 杂务

题解：递推关系

 数据：

input:

10
1 3 0
2 9 1 0
3 2 1 2 0
4 2 1 2 3 0
5 12 2 0
6 1 1 2 3 4 5 0
7 17 3 0
8 7 1 2 3 4 5 6 7 0
9 13 1 3 8 0
10 2 1 2 3 4 5 6 7 8 9 0

output:

53

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<stack>
using namespace std;
int n,cnt,at[10005],ru[10005],hd[10005],ans, x,t,y, f[10005];
stack<int> q;
struct Edge{
    int to, nxt;
}edge[1000005];
void add(int u, int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
int main(){
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        scanf("%d%d",&x,&t);
        at[x] = t;
        while(~scanf("%d",&y)){
            if(y == 0) break;
            add(y,x);
            ru[x]++;
        }
    }
    int mx = 0;
    for(int i = 1; i <= n; i++)
        if(ru[i] == 0){
            q.push(i);
            f[i] = at[i];
        }
    while(!q.empty()){
        int u = q.top(); q.pop();
        for(int i = hd[u]; i; i = edge[i].nxt){
            int v = edge[i].to;
            ru[v]--;
            if(!ru[v]){
                q.push(v);
            }
            f[v] = max(f[v], f[u] + at[v]);
            ans = max(ans, f[v]);
        }
    }
    printf("%d\n",ans);
    return 0;
}

 

P1983 车站分级

 P3074 [USACO13FEB]Milk Scheduling S

小X的N头奶牛要挤奶N∈[1,10000].为了方便编号1…N.每头奶牛挤奶耗时T[i] .但是有些奶牛一定要在另一头奶牛前挤奶。为了加快挤奶速度，小X雇佣了无限个农场工人。可以同时挤奶，求最少的时间挤完所有奶。

ｕ１－ｖ

ｕ２－ｖ

ｄｉｓ［ｖ］　＝　ｍａｘ（ｄｉｓ［ｕ］＋　ｄｉｓ［ｖ］（最开始的ｄｉｓ［ｖ］，否则两者会叠加，因此使用ｐｒｅ数组））

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+5;
int n, m, ru[N], chu[N], x, y, ans, hd[N], dis[N], cnt, pre[N];
queue<int> q;
struct Edge{
	int to, nxt;
}edge[50005];
void add(int u, int v){
	cnt++;
	edge[cnt].nxt = hd[u];
	edge[cnt].to = v;
	hd[u] = cnt;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= n; i++){
		scanf("%d",&dis[i]); pre[i] = dis[i];
	}
	for(int i = 1; i <= m; i++){
		scanf("%d%d",&x,&y);
		add(x,y); ru[y]++; chu[x]++;
	}
	for(int i = 1; i <= n; i++){
		if(!ru[i]){
			q.push(i);
		}
	}
	while(!q.empty()){
		int u = q.front(); q.pop();
		if(!chu[u]){
			ans = max(ans, dis[u]);
		}
		for(int i = hd[u]; i; i = edge[i].nxt){
			int v = edge[i].to; ru[v]--;
			dis[v] = max(dis[v], dis[u] + pre[v]);
//			cout<<"v: "<<v<<" "<<dis[v]<<endl;
			if(ru[v] == 0)	
//				dis[v] = max(dis[v], dis[u] + dis[v]), 
				q.push(v);
		}
	}
	printf("%d\n",ans);
	return 0;
} 
/**
10 10
1 2 3 4 5 6 7 8 9 10
3 4
5 4
5 10
4 6
4 10
10 1
6 1
6 2
1 8
9 7
*/