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  • UVa 11090

    题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=2031&mosmsg=Submission+received+with+ID+26584370

    求有向图平均权最小的回路

    二分答案 (x),将所有边都减去 (x),看图中是否有负环即可

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    const int maxn = 105;
    const double eps = 1e-3;
    
    int T, n, m; 
    
    int h[maxn], cnt = 0;
    struct E{
    	int to, next;
    	double cost;
    }e[maxn * maxn];
    void add(int u, int v, double w){
    	e[++cnt].to = v;
    	e[cnt].cost = w;
    	e[cnt].next = h[u];
    	h[u] = cnt;
    }
    
    int in[maxn], tot[maxn]; double d[maxn];
    bool spfa(double mid){
    	queue<int> q;
    	memset(tot, 0, sizeof(tot));
    	memset(in, 0, sizeof(in));
    	for(int i = 1 ; i <= n ; ++i) {
    		d[i] = 0;
    		in[i] = 1;
    		q.push(i);
    	}
    	
    	while(!q.empty()){
    		int u = q.front(); q.pop(); in[u] = 0;
    		for(int i = h[u] ; i != -1 ; i = e[i].next){
    			int v = e[i].to;
    			if(d[u] + e[i].cost - mid - d[v] < 0){
    				d[v] = d[u] + e[i].cost - mid;
    				
    				if(!in[v]){
    					in[v] = 1;
    					q.push(v);
    					++tot[v];
    					if(tot[v] > n) return false;
    				}
    			}
    		}
    	}
    	
    	return true;
    }
    
    bool check(double x){
    	return !spfa(x);
    }
    
    ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }
    
    int main(){
    	scanf("%d", &T);
    	int kase = 0;
    	while(T--){
    		memset(h, -1, sizeof(h)); cnt = 0;
    		scanf("%d%d", &n, &m);
    		int u, v; double w;
    		double L = 0, R = 0;
    		for(int i = 1 ; i <= m ; ++i){
    			scanf("%d%d%lf", &u, &v, &w);
    			add(u, v, w);
    			R = max(R, w * 1.0);
    		}
    		
    		printf("Case #%d: ", ++kase);	
    		if(!check(R + 1.0)) printf("No cycle found.
    ");
    		else {
    			double ans = 0;
    			while(R - L > eps){
    				double mid = (R + L) / 2;
    				if(check(mid)){
    					R = mid;
    				} else{
    					L = mid;
    				}
    			}
    			printf("%.2lf
    ", L);
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/tuchen/p/15026889.html
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