• UVa 11478


    题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=2473&mosmsg=Submission+received+with+ID+26584758

    题目条件可以转化为每个点最终的权值为:(w + sum[u] - sum[v]),二分答案,判断最小的边可不可以是 (x),则有 (sum[v] <= sum[u] + w - x),即转化为差分约束系统,若系统中有负环,则无解(大小关系矛盾)

    如果要求方案,添加一个源点 (S)(S) 向其他所有点连边,边权为 (0),跑一遍 (spfa),则从源点到每个点的距离就是答案

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    const int maxn = 510;
    const int maxm = 3010;
    
    int n, m;
    
    int h[maxn], cnt = 0;
    struct E{
    	int to, cost, next;
    }e[maxm];
    void add(int u, int v, int w){
    	e[++cnt].to = v;
    	e[cnt].cost = w;
    	e[cnt].next = h[u];
    	h[u] = cnt;
    }
    
    int in[maxn], tot[maxn], d[maxn];
    bool spfa(int 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] ){
    				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(int 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(){
    	while(scanf("%d%d", &n, &m) != EOF){
    		memset(h, -1, sizeof(h)); cnt = 0;
    		int u, v, w;
    		
    		int L = 0, R = 0;
    		for(int i = 1 ; i <= m ; ++i){
    			scanf("%d%d%d", &u, &v, &w);
    			add(u, v, w);
    			R = max(R, w);
    		}
    		
    		if(check(R + 1)) printf("Infinite
    ");
    		else if(!check(1)) printf("No Solution
    ");
    		else {
    			int ans = 0;
    			while(L <= R){
    				int mid = (L + R) >> 1;
    				if(check(mid)){
    					L = mid + 1;
    					ans = mid;
    				} else{
    					R = mid - 1;
    				}
    			}
    			printf("%d
    ", ans);			
    		}
    
    	}
    	return 0;
    }
    
  • 相关阅读:
    centos7 & ubuntu14.02安装sublime 3
    flask之flask-restful
    ubuntu14.04安装python3.7.1
    vim中多行注释和多行删除命令
    python3之scrapy安装使用
    python3 之 linux命令实现
    ubuntu14.04安装pyspider
    升级3.4成3.6 ubuntu14.04 和miniconda虚拟环境
    python3 之初学者常犯的5个错误
    python3 之 格式化json
  • 原文地址:https://www.cnblogs.com/tuchen/p/15027291.html
Copyright © 2020-2023  润新知