hdu4786 Fibonacci Tree（2013 成都 ），确定上下界，生成树

思维题

确定上下界，1边最少l个，最多r个，根据树的构造过程，必定可以用1边替换0边从而可到达1边l到r的个数

参考：http://blog.youkuaiyun.com/kk303/article/details/16368165

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FD(i, b, a) for(int i = (b); i >= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(a, v) memset(a, v, sizeof(a))
#define PB push_back
#define MP make_pair

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MAXN = 100010;

int n, m;
struct Edge{
    int from, to, l;
    Edge(){}
    Edge(int from, int to, int l) : from(from), to(to), l(l) {}
    bool operator<(const Edge& rhs) const
    {
        return l < rhs.l;///升序
    }
}E[MAXN];

int fa[MAXN];
void init_fa(int n)
{
    for (int i = 0; i <= n; i++) fa[i] = i;
}
int findset(int x){ return x == fa[x] ? x : fa[x] = findset(fa[x]); }

int solve(int op)
{
    int l, addi;
    if (op) l = m - 1, addi = -1;
    else l = 0, addi = 1;

    init_fa(n);
    int ans = 0;
    int cnt = 0;
    for (int i = l; i <= m - 1 && i >= 0; i += addi)
    {
        int x = E[i].from, y = E[i].to, l = E[i].l;
        int fax = findset(x), fay = findset(y);
        if (fax != fay)
        {
            fa[fax] = fay;
            ans += E[i].l;
            cnt++;
            if (cnt >= n - 1) break;
        }
    }
    if (cnt < n - 1) return -1;
    else return ans;
}

int fb[MAXN];
int fb_cnt;
void pre()
{
    fb[0] = 1, fb[1] = 2;
    int i;
    for (i = 2; ;i++)
    {
        fb[i] = fb[i - 1] + fb[i - 2];
        if (fb[i] >= 100000) break;
    }
    fb_cnt = i;
}

int main ()
{
    int nc = 1;
    int T;
    pre();
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d",&n, &m);
        REP(i, m)
        {
            scanf("%d%d%d", &E[i].from, &E[i].to, &E[i].l);
        }
        printf("Case #%d: ", nc++);
        sort(E, E + m);
        int lm , rm;
        lm = solve(0);
        if (lm < 0)
        {
            printf("No\n");
            continue;
        }
        rm = solve(1);
        int fla = 0;
        for (int i = 0; i <= fb_cnt; i++)
        {
            if (fb[i] <= rm && fb[i] >= lm)
            {
                fla = 1;
                break;
            }
        }
        if (fla) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}