Traveler Nobita （zoj 3456 最小生成树）

Traveler Nobita

Time Limit: 2 Seconds
Memory Limit: 65536 KB

One day, Nobita used a time machine and went back to 1000 AD. He found that there are
N cities in the kingdom he lived. The cities are numbered from 0 to
N - 1. Before 1000 AD., there are no roads between any two cities. The kingdom will build one road between two cities at the beginning of each year starting from 1000 AD. There might be duplicated roads between two cities being built by the kingdom. You
can assume that building a road takes no time.

At the beginning of every year, after the new road is built, Nobita will try to make a schedule to travel around all cities within that year. The travel should both begin at and end at the capital city - city₀. Every time Nobita arrived at a city
i, he will spent t1_i days in that city, regardless of how many times he had come to the city. Of course he wouldn't need to spend any time in the capital city (that is to say,
t1₀ is always 0). And t2_i hours is required to pass road #i. Note that to pass each road, a passport of that road is required. And the kingdom limits that one person can only have no more than
N - 1 passports of roads each year.

You are given information about the roads built in M years. Please find out the minimum time Nobita needed to complete his traveling schedule.

Input

There are multiple cases. The first line of a test case contains two integers,
N (2 ≤ N ≤ 200) and M (1 ≤ M ≤ 10000). The next line contains
N integers, indicating t1₀ ... t1_{n - 1}. (0 ≤
t1_i ≤ 50) The next M lines, the i^th (0 ≤
i < M) line of this section contains three integers, u_i,
v_i, t2_i, (0 ≤ u_i,
v_i < N; 0 ≤ t2_i ≤ 5000), indicating that in year
1000 + i AD., a road will be built between city u_i and city
v_i. t1_i and t2_i have been described above.

Output

For each case, you should output M lines. For the i^th line, if Nobita can make a schedule in year
1000 + i, output the minimal days he can finish that schedule, rounded to two decimal digits. Otherwise output -1. There should be a blank line after each case.

Sample Input

Sample Output

题意：n个点m条路，開始没有路。每一年修一条路。修完后一个人从0点周游这n个点。问是否能在一年内游玩这n个点，能的话输出最少的天数。输入会告诉每一个点他待的时间和每条路走的时间，他最多仅仅能走n-1条路。

思路：一边加边一边Kruskal，每次Kruskal把没实用的边删掉，另外前n-2年肯定不能完毕。还要注意闰年。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <vector>

#include <set>

#include <queue>

#pragma comment (linker,"/STACK:102400000,102400000")

#define maxn 2005

#define MAXN 20025

#define mod 1000000009

#define INF 0x3f3f3f3f

#define pi acos(-1.0)

#define eps 1e-6

#define lson rt<<1,l,mid

#define rson rt<<1|1,mid+1,r

#define FRE(i,a,b)  for(i = a; i <= b; i++)

#define FREE(i,a,b) for(i = a; i >= b; i--)

#define FRL(i,a,b)  for(i = a; i < b; i++)

#define FRLL(i,a,b) for(i = a; i > b; i--)

#define mem(t, v)   memset ((t) , v, sizeof(t))

#define sf(n)       scanf("%d", &n)

#define sff(a,b)    scanf("%d %d", &a, &b)

#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)

#define pf          printf

#define DBG         pf("Hi\n")

typedef long long ll;

using namespace std;

struct Edge

{

    int u,v,len;

    bool operator<(const Edge &a)const

    {

        return len<a.len;

    }

};

vector<Edge>edge;

int father[maxn];

int a[maxn];

bool vis[maxn];

int num=0;

int n,m;

void init()

{

    num=0;

    edge.clear();

}

void addedge(int u,int v,int w,int id)

{

    Edge e={u,v,(a[u]+a[v])*24+w*2};

    edge.push_back(e);

}

int find_father(int  x)

{

    if (x!=father[x])

        father[x]=find_father(father[x]);

    return father[x];

}

int Kruskal()

{

    int i,j;

    for (i=0;i<n;i++) father[i]=i;

    sort(edge.begin(),edge.end());

    int cnt=0,ans=0;

    for (vector<Edge>::iterator it = edge.begin();it!=edge.end();)

    {

        int u=it->u;

        int v=it->v;

        int l=it->len;

        int fu=find_father(u);

        int fv=find_father(v);

        if (fu!=fv)

        {

            ans+=l;

            it++;

            cnt++;

            father[fu]=fv;

        }

        else

            edge.erase(it);

//        if (cnt==n-1) break;      //不要break。要把后面无关的边删掉，不然sort会耗时

    }

    if (cnt<n-1) return -1;

    return ans;

}

bool isok(int x)

{

    if ((x%4==0&&x%100)||x%400==0) return true;

    return false;

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);

#endif

    int i,j;

    while (~sff(n,m))

    {

        init();

        for (i=0;i<n;i++)

            sf(a[i]);

        int u,v,w;

        for (i=0;i<m;i++)

        {

            sfff(u,v,w);

            addedge(u,v,w,i);

            int x=Kruskal();

            if (x==-1) {

                pf("-1\n");

                continue;

            }

            int yy;

            if (isok(1000+i)) yy=366;

            else yy=365;

            if (yy*24<x){

                pf("-1\n");

                continue;

            }

            pf("%.2lf\n",x/24.0);

        }

        pf("\n");

    }

    return 0;

}