标签:des style blog http java color
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16075 Accepted Submission(s): 6677
3 3 1 2 1 1 3 2 2 3 4 1 3 2 3 2 0 100
3 ?
这是一道带权值的并查集,因为是中文题, 题意也就不用多说了。。
所以我是用结构体写的。。。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<queue> #include<sstream> #include<cmath> using namespace std; #define f1(i, n) for(int i=0; i<n; i++) #define f2(i, n) for(int i=1; i<=n; i++) #define f3(i, n) for(int i=n; i>=1; i--) #define f4(i, n) for(int i=1; i<n; i++) #define M 10050 int f[M]; int r[M]; int ans; int t; int n, m; int coun; struct node { int x; int y; int cost; //花费 }q[M]; int cmp(node x1, node y1) { return x1.cost < y1.cost; } int find(int x) //并查集的find { return f[x] == x ? x:f[x] = find( f[x] ); } void Kruskal() { sort(q, q+n, cmp); f2(i, n) { int xx = find(q[i].x); int yy = find(q[i].y); if( xx!=yy ) //当不是统一集合时。。 { ans+=q[i].cost; f[yy] = xx; coun --; //连起来一条路 } } } int main() { while(scanf("%d%d", &n, &m) &&n) { ans = 0; t = 0; coun = m; f2(i, m) f[i] = i; //初始化 f2(i, n) scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].cost); Kruskal(); if(coun==1) printf("%d\n", ans); else printf("?\n"); } return 0; }
HDU 1863:畅通工程(带权值的并查集),布布扣,bubuko.com
标签:des style blog http java color
原文地址:http://blog.csdn.net/u013487051/article/details/37805331