Reward

2 1
1 2
2 2
1 2
2 1

1777
-1

题意：老板要给n个员工发年终奖金，然后给出m个需求 (a,b)，表示a的奖金要比b的多，然后呢，老板决定每个员工至少可以拿888元，现在老板总共至少需要发多少奖金。

            如果满足不了就输出-1。

分析：拓扑排序问题。比第i个人奖金少的总人数有sum个，那么第i个人的奖金就是888+sum元。注意此题数据比较大，用邻接表存吧。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2647

代码清单：

(1) vector实现

#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;

const int maxv = 10000 +5;

int n,m;
int p,q;
bool judge;
int save[maxv];
int degree[maxv];
vector<int>graph[maxv];

void init(){
    memset(degree,0,sizeof(degree));
    for(int i=0;i<=n;i++) graph[i].clear();
}

int topSort(){
    int cnt=0;
    int sum=0;
    int q=888;
    while(cnt<n){
        int flag=0;
        for(int i=1;i<=n;i++){
            if(degree[i]==0){
                degree[i]=-1;
                save[flag++]=i;
                cnt++;
            }
        }
        if(!flag) return -1;
        sum+=q*flag; q++;
        for(int i=0;i<flag;i++){
            int u=save[i];
            for(int j=0;j<graph[u].size();j++){
                int v=graph[u][j];
                degree[v]--;
            }
        }
    }return sum;
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=0;i<m;i++){
            scanf("%d%d",&p,&q);
            judge=false;
            for(int j=0;j<graph[q].size();j++){
                if(graph[q][j]==p){
                    judge=true;
                    break;
                }
            }
            if(!judge){
                graph[q].push_back(p);
                degree[p]++;
            }
        }printf("%d\n",topSort());
    }return 0;
}

(2)静态邻接表(数组)

#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;

const int maxv = 10000 + 5;
const int maxn = 20000 + 5;

struct Edge{
    int to,next;
}graph[maxn];

int n,m;
int p,q;
int index;
bool judge;
int save[maxv];
int head[maxn];
int degree[maxv];

void init(){
    index=1;
    memset(head,0,sizeof(head));
    memset(degree,0,sizeof(degree));
}

void add(int u,int v){
    graph[index].to=v;
    graph[index].next=head[u];
    head[u]=index++;
}

int topSort(){
    int cnt=0;
    int sum=0;
    int q=888;
    while(cnt<n){
        int flag=0;
        for(int i=1;i<=n;i++){
            if(degree[i]==0){
                degree[i]=-1;
                save[flag++]=i;
                cnt++;
            }
        }
        if(!flag) return -1;
        sum+=q*flag; q++;
        for(int i=0;i<flag;i++){
            int u=save[i];
            for(int j=head[u];j!=0;j=graph[j].next){
                int v=graph[j].to;
                degree[v]--;
            }
        }
    }return sum;
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=0;i<m;i++){
            scanf("%d%d",&p,&q);
            judge=false;
            for(int j=head[q];j!=0;j=graph[j].next){
                if(graph[j].to==p){
                    judge=true;
                    break;
                }
            }
            if(!judge){
                add(q,p);
                degree[p]++;
            }
        }printf("%d\n",topSort());
    }return 0;
}

HDU_2647_Reward(拓扑排序)

原文：http://blog.csdn.net/jhgkjhg_ugtdk77/article/details/45566293