A - Promotions

题目详见http://7xjob4.com1.z0.glb.clouddn.com/3f644de6844d64706eb36baa3a0c27b0

这题是普通的拓扑排序，要把每一层的都保存下来。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define MOD 1000000007
struct edge{
    int to,next;
}e[2*20050];
int first[5050],du[5050];
int q[1111111];
int n,m;
int ceng[5050][5050],num[5050],deep[5050],cengshu,sum[5050];

void topu()
{
    int i,j;
    int front,rear;
    front=1;rear=0;
    memset(num,0,sizeof(num));
    memset(deep,0,sizeof(deep));
    for(i=1;i<=n;i++){
        if(!du[i]){
            rear++;
            q[rear]=i;
            deep[i]=1;
        }
    }
    int x,y,xx,yy;
    cengshu=1;
    while(front<=rear){
        x=q[front];
        front++;
        cengshu=max(cengshu,deep[x]);
        num[cengshu]++;
        ceng[cengshu ][num[cengshu] ]=x;
        for(i=first[x];i!=-1;i=e[i].next){
            int v=e[i].to;
            du[v]--;
            if(!du[v]){
                rear++;
                q[rear]=v;
                deep[v]=deep[x]+1;
            }
        }
    }

    sum[0]=0;
    for(i=1;i<=cengshu;i++){
        sum[i]=sum[i-1]+num[i];
    }
}

int main()
{
    int i,j,l,r,tot,c,d;
    while(scanf("%d%d%d%d",&l,&r,&n,&m)!=EOF)
    {
        memset(first,-1,sizeof(first));
        memset(du,0,sizeof(du));
        tot=0;
        for(i=1;i<=m;i++){
            scanf("%d%d",&c,&d);
            c++;d++;
            du[d]++;
            tot++;
            e[tot].next=first[c];e[tot].to=d;
            first[c]=tot;
        }

        topu();
        int num1,num2,num3;
        for(i=1;i<=cengshu;i++){
            if(sum[i]>l)break;
        }
        i--;
        num1=sum[i];

        for(i=1;i<=cengshu;i++){
            if(sum[i]>r)break;
        }
        i--;
        num2=sum[i];

        for(i=1;i<=cengshu;i++){
            if(sum[i]>=r)break;
        }
        if(i>=cengshu)num3=0;
        else num3=sum[cengshu]-sum[i];
        printf("%d\n%d\n%d\n",num1,num2,num3);
    }
    return 0;
}