标签:mit mes else print == namespace dig typename limits
题目链接:https://vjudge.net/problem/POJ-2377#author=tsacm123
题目大意:就是让你算出n个谷仓之间的最大生成树,然后把各条边的值累加起来
#include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl ‘\n‘ #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) fill(a, a+maxn, INF); #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; typedef pair<double, int> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == ‘-‘)f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 1e3+10; int vis[maxn]; vector<P> dis[maxn]; void prim(int n) { priority_queue<P> pq; zero(vis); int sum = 0, s = 0; pq.push(make_pair(0, 1)); while(!pq.empty()) { P t = pq.top(); pq.pop(); if (vis[t.second]) continue; vis[t.second] = true; sum += t.first; ++s; for (int i = 0; i<(int)dis[t.second].size(); ++i) if (!vis[dis[t.second][i].second]) pq.push(make_pair(dis[t.second][i].first, dis[t.second][i].second)); } if (s != n) printf("-1\n"); else printf("%d\n", sum); } int main(void) { int n, m; scanf("%d%d", &n, &m); for (int i = 0, a, b, d; i<m; ++i) { scanf("%d%d%d", &a, &b, &d); dis[a].push_back(make_pair(d, b)); dis[b].push_back(make_pair(d, a)); } prim(n); return 0; }
POJ - 2377 - Bad Cowtractors (最小生成树)
标签:mit mes else print == namespace dig typename limits
原文地址:https://www.cnblogs.com/shuitiangong/p/12384759.html