链接:
http://acm.hdu.edu.cn/showproblem.php?pid=3938
题目:
Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.
Input
There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).
Output
Output the answer to each query on a separate line.
Sample Input
10 10 10 7 2 1 6 8 3 4 5 8 5 8 2 2 8 9 6 4 5 2 1 5 8 10 5 7 3 7 7 8 8 10 6 1 5 9 1 8 2 7 6
Sample Output
36 13 1 13 36 1 36 2 16 13
分析与总结:
做这题学到了什么是“离线算法”的概念。所谓“离线”,就是把所有的数据都输入之后再计算,“在线”就是边输入边计算。
用在这题中,是因为输入中的“询问部分”,有Q 个问,每个L可以有多少种不同路径。由于大的L必定会包含到小的L, 所以把所有问题都输入,再从大到小排序,再计算,可以减少很多计算量。
本栏目更多精彩内容:http://www.bianceng.cn/Programming/sjjg/
这题还需要用到的是并查集中的“权值”, 用rank数组表示,也就是某个棵树k有rank【k】个结点。同一个树之间的点都是连通的,任何点都可以通往其它的任意点, 那么当两颗树合并成一棵树时, 将会增加rank[a]*rank[b]条路径。
代码:
#include<cstdio> #include<algorithm> using namespace std; #define N 10005 int f[N], rank[N], ans[N], n, m, Q; struct Edge{ int u, v, val; friend bool operator < (const Edge&a,const Edge&b){ return a.val < b.val; } }arr[N*5]; struct Query{ int id, L; friend bool operator<(const Query&a,const Query&b){ return a.L<b.L; } }q[N]; void init(){ for(int i=0; i<=n; ++i) f[i]=i, rank[i]=1; } int find(int x){ int i, j=x; while(j!=f[j]) j=f[j]; while(x!=j){ i=f[x]; f[x]=j; x=i; } return j; } int Union(int x,int y){ int a=find(x), b=find(y); if(a==b) return 0; int t=rank[a]*rank[b]; rank[a] += rank[b]; f[b] = a; return t; } int main(){ while(~scanf("%d%d%d",&n,&m,&Q)){ for(int i=0; i<m; ++i) scanf("%d%d%d",&arr[i].u,&arr[i].v,&arr[i].val); for(int i=0; i<Q; ++i) scanf("%d",&q[i].L), q[i].id=i; sort(arr,arr+m); sort(q,q+Q); int cnt=0, j=0; init(); for(int i=0; i<Q; ++i){ while(j<m && arr[j].val<=q[i].L){ cnt += Union(arr[j].u, arr[j].v); ++j; } ans[q[i].id] = cnt; } for(int i=0; i<Q; ++i) printf("%d\n", ans[i]); } return 0; }
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, 这题怎么做啊
, hdu1716 排序 oj
, 输入
, 10
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