标签：数据 nbsp output 16px main names operator 不用 bsp
【数据范围】
对于30%的数据，保证 1<=N<=2000
对于100%的数据，保证 1<=N<=100000
对于所有的数据，保证 1<=M<=1000000，1<=Hi<=1000000000，1<=Ki<=1000000000。
对于第一问，我们直接遍历一遍就好了。
对于第二问，把第一问中的边取出。结点具有层次性，且不具有环。把边先按终点高度排序为第一关键字(从大到小)，边长为第二关键字排序(从大到小)之后，就会保证优先到高点,同高点之间选最小边。
1 //It is made by Awson on 2017.10.26 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <cmath> 7 #include <stack> 8 #include <queue> 9 #include <vector> 10 #include <string> 11 #include <cstdio> 12 #include <cstdlib> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 #define LL long long 17 #define link LINK 18 #define Min(a, b) ((a) < (b) ? (a) : (b)) 19 #define Max(a, b) ((a) > (b) ? (a) : (b)) 20 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 21 using namespace std; 22 const int N = 1e5; 23 const int M = 1e6; 24 25 int n, m, h[N+5], u, v, w, ans1; 26 LL ans2; 27 struct tt { 28 int to, cost, next; 29 }edge[(M<<1)+5]; 30 struct ss { 31 int from, to, cost; 32 bool operator < (const ss &b) const{ 33 return h[to] == h[b.to] ? cost < b.cost : h[to] > h[b.to]; 34 } 35 }link[(M<<1)+5]; 36 int path[N+5], top; 37 bool vis[N+5]; 38 int st[N+5]; 39 40 int find(int r) { 41 return st[r] ? st[r] = find(st[r]) : r; 42 } 43 void bfs() { 44 queue<int>Q; 45 while (!Q.empty()) Q.pop(); 46 Q.push(1); vis[1] = 1; 47 while (!Q.empty()) { 48 int u = Q.front(); ans1++; Q.pop(); 49 for (int i = path[u]; i; i = edge[i].next) 50 if (!vis[edge[i].to]) { 51 Q.push(edge[i].to); vis[edge[i].to] = 1; 52 } 53 } 54 } 55 void add(int u, int v, int c) { 56 edge[++top].to = v; 57 edge[top].cost = c; 58 edge[top].next = path[u]; 59 path[u] = top; 60 } 61 void Kruskal() { 62 int cnt = 0; 63 for (int u = 1; u <= n; u++) 64 for (int j = path[u]; j; j = edge[j].next) 65 if (vis[u] && vis[edge[j].to]) 66 link[++cnt].from = u, link[cnt].to = edge[j].to, link[cnt].cost = edge[j].cost; 67 sort(link+1, link+1+cnt); 68 for (int i = 1; i <= cnt; i++) { 69 int u = link[i].from, v = link[i].to, c = link[i].cost; 70 int p = find(u), q = find(v); 71 if (p != q) { 72 st[p] = q; ans2 += c; 73 } 74 } 75 } 76 void work() { 77 scanf("%d%d", &n, &m); 78 for (int i = 1; i <= n; i++) scanf("%d", &h[i]); 79 for (int i = 1; i <= m; i++) { 80 scanf("%d%d%d", &u, &v, &w); 81 if (h[u] >= h[v]) add(u, v, w); 82 if (h[v] >= h[u]) add(v, u, w); 83 } 84 bfs(); 85 Kruskal(); 86 printf("%d %lld\n", ans1, ans2); 87 } 88 int main() { 89 work(); 90 return 0; 91 }
标签：数据 nbsp output 16px main names operator 不用 bsp
原文地址：http://www.cnblogs.com/NaVi-Awson/p/7739434.html
