InputOne 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.OutputFor 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 1
Sample Output
1777 -1
题意：输入两个数n,m。n表示公司人数，m表示不同人之间的关系。a b表示a的工资大于b.
题解：拓扑排序，找入度为0的点，所有的拓扑完没有剩余，说明没有环
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<stack> #include<map> #include<cstdlib> #include<vector> #include<string> #include<queue> using namespace std; #define ll long long #define llu unsigned long long #define INF 0x3f3f3f3f const double PI = acos(-1.0); const int maxn = 1e4+10; const int mod = 1e9+7; int degree[maxn],add[maxn]; vector<int>G[maxn]; int n,m; int toposort() { queue<int>que; for(int i=1;i<=n;i++) if(degree[i] == 0) que.push(i); int ans = 0; while(que.size()) { int ptr = que.front(); que.pop(); ans++; for(int i=0;i<G[ptr].size();i++) { degree[G[ptr][i]]--; if(degree[G[ptr][i]] == 0) { que.push(G[ptr][i]); add[G[ptr][i]] = add[ptr] + 1; } } } if(ans != n) return -1; int sum = 0; for(int i=1;i<=n;i++) sum += add[i] + 888; return sum; } int main() { while(~scanf("%d%d",&n,&m)) { memset(degree,0,sizeof degree); memset(add,0,sizeof add); for(int i=1;i<=n;i++) G[i].clear(); for(int i=0;i<m;i++) { int a,b; scanf("%d%d",&a,&b); G[b].push_back(a); degree[a]++; } printf("%d\n",toposort()); } }
原文：https://www.cnblogs.com/smallhester/p/10389783.html