多校赛2017 R1
#多校赛2017#
HDU 6044
Problem
As to a permutation p1,p2,⋯,pn from 1 to n, it is uncomplicated for each 1≤i≤n to calculate (li,ri) meeting the condition that min(pL,pL+1,⋯,pR)=pi if and only if li≤L≤i≤R≤ri for each 1≤L≤R≤n.
Given the positive integers n, (li,ri) (1≤i≤n), you are asked to calculate the number of possible permutations p1,p2,⋯,pn from 1 to n, meeting the above condition.
The answer may be very large, so you only need to give the value of answer modulo 109+7.
题意
给n个L,R对，对于第i个对，表示其在(al, a(l+1),…ar)中为最小值，问有多少种（1…n）的置换满足条件。
思路
不考虑数字的组合方式，对于给定的LR对生成的二叉树唯一，直接搜索即可。L升序R降序快排后发现，此时LR序列正好为二叉树的先序遍历结果！（贼关键）
#includeHDU 6040
Problem
sd0061, the legend of Beihang University ACM-ICPC Team, retired last year leaving a group of noobs. Noobs have no idea how to deal with m coming contests. sd0061 has left a set of hints for them.
There are n noobs in the team, the i-th of which has a rating ai. sd0061 prepares one hint for each contest. The hint for the j-th contest is a number bj, which means that the noob with the (bj+1)-th lowest rating is ordained by sd0061 for the j-th contest.
The coach asks constroy to make a list of contestants. constroy looks into these hints and finds out: bi+bj≤bk is satisfied if bi≠bj, bi
题意
给1e7个数，有100次查询，每次求第bi+1小的数，bi相当于斐波那契数列，这样就可以做到每次查询的范围缩小1.618倍，然后复杂度就是Sigma(n/1.618^i),其实就是求个等比数列。
思路
会求复杂度就敢搞了。
#includeHDU 6041
Problem
There is a connected undirected graph with weights on its edges. It is guaranteed that each edge appears in at most one simple cycle.
Assuming that the weight of a weighted spanning tree is the sum of weights on its edges, define V(k) as the weight of the k-th smallest weighted spanning tree of this graph, however, V(k) would be defined as zero if there did not exist k different weighted spanning trees.
Please calculate (∑k=1Kk⋅V(k))mod232.
题意
仙人掌图，求1到k小的最小生成树权值和。
思路
点双连通优先队列暴搞看脸A。
#include pq;
for(int i = 0; i < vb.size(); i++) {
pq.push((nd) {va[0] + vb[i], 0, i});
}
sv.resize(0);
while(sv.size() < K && !pq.empty()) {
nd nt = pq.top(); pq.pop();
sv.push_back(nt.w);
if(nt.x + 1 < va.size()) {
nt.x ++;
pq.push((nd) {va[nt.x] + vb[nt.y], nt.x, nt.y});
}
}
va = sv;
}
const int MAXN = 1010;
const int MAXM = 5000;
struct Edge
{
int to,next,val;
} edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN];//,Stack[MAXN],Belong[MAXN];
int Index,top;
int block;//点双连通分量的个数
//bool Instack[MAXN];
void addedge(int u,int v,int val)
{
edge[tot].to = v;edge[tot].next = head[u];edge[tot].val = val;head[u] = tot++;
}
stack<int> st;
void Tarjan(int u,int pre = -1)
{
int v;
Low[u] = DFN[u] = ++Index;
//Stack[top++] = u;
//Instack[u] = true;
for(int i = head[u];i != -1;i = edge[i].next)
{
v = edge[i].to;
//if(v == pre)continue;
if( !DFN[v] )
{
st.push(i);
Tarjan(v,u);
if(Low[u] > Low[v])Low[u] = Low[v];
if( Low[v] >= DFN[u])
{
block++;
int ii;
vector<int> t2;
do
{
//vn = Stack[--top];
//Belong[vn] = block;
//Instack[vn] = false;
ii = st.top();
st.pop();
t2.push_back(edge[ii].val);
}
while( ii!=i );
// cout << t2.size() << endl;
// for (auto &x : t2) cout << x << ' '; cout << endl;
if(t2.size() > 1) Solve(res, t2);
}
}
else if(v!=pre && DFN[v] < DFN[u]) {
st.push(i);
Low[u] = min(Low[u], DFN[v]);
}
// else if(Instack[v] && Low[u] > DFN[v])
// {
// Low[u] = DFN[v];
// st.push(i);
// }
}
}
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
int main(){
//freopen("1009.in","r",stdin);
int n, m, cs = 1;
res.reserve(100005);
sv.reserve(100005);
while(~scanf("%d%d",&n,&m)) {
init();
int x, y, z;
int sum = 0;
for(int i = 1; i <= m; i++) {
scanf("%d%d%d",&x,&y,&z);
addedge(x,y,z);
addedge(y,x,z);
sum += z;
}
res.resize(0);
res.push_back(0);
scanf("%d",&K);
memset(DFN,0,sizeof(DFN));
//memset(Instack,false,sizeof(Instack));
//memset(Belong,0,sizeof(Belong));
Index = block = top = 0;
Tarjan(1);
unsigned r = 0;
for(int i = 0; i < res.size(); i++) {
r += (unsigned)(i + 1) * (unsigned)(sum - res[i]);
}
printf("Case #%d: %u\n",cs++,r);
//if(n == 28) return 0;
}
}