题目 Reward
Dandelion’s uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a’s reward should more than b’s.Dandelion’s unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work’s reward will be at least 888 , because it’s a lucky number.
Input
One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a’s reward should be more than b’s.
Output
For every case ,print the least money dandelion ‘s uncle needs to distribute .If it’s impossible to fulfill all the works’ demands ,print -1.
Sample Input
2 1
1 2
2 2
1 2
2 1Sample Output
1777
-1
注意初始化~~
#include<iostream>
#include<cstdio> //EOF,NULL
#include<cstring> //memset
#include<cstdlib> //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc
#include<cmath> //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include<algorithm> //fill,reverse,next_permutation,__gcd,
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iterator>
#include<iomanip> //setw(set_min_width),setfill(char),setprecision(n),fixed,
#include<functional>
#include<map>
#include<set>
#include<limits.h> //INT_MAX
#include<bitset> // bitset<?> n
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define pb(x) push_back(x)
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
const int INF =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 10010;
int n,m;
int a,b;
int in[MAXN];
int sum,cnt;
int price[MAXN];
vector<int>Edge[MAXN];
vector<int> ans;
queue<int> q;
void Init(){
for(int i = 1;i <= n;i++){
Edge[i].clear();
}
memset(in,0,sizeof in);
while(!q.empty()) q.pop();
sum = cnt = 0;
}
int main(){
while(read2(n,m)!=EOF){
Init();
for(int i = 0 ; i < m;i++){
read2(a,b);
Edge[b].push_back(a); //反向建图
in[a] ++;
}
for(int i = 1 ;i <= n ;i++){
if(in[i] == 0){
q.push(i);
price[i] = 888;
}
}
while(!q.empty()){
int p = q.front();
sum += price[p];
cnt++;
q.pop();
for(int i = 0; i < Edge[p].size(); i++){
int y = Edge[p][i];
in[y] --;
price[y] = price[p]+1;
if(in[y] == 0){
q.push(y);
}
}
}
if(cnt < n){
printf("-1\n");
}
else{
print(sum);
}
}
}
转载于:https://www.cnblogs.com/llke/p/10780133.html
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/101021.html原文链接：https://javaforall.net
