A - Promotions

来源：互联网发布：华为视频会议软件编辑：程序博客网时间：2024/05/02 14:11

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 1000000007struct 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;}

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