P2323 [HNOI2006]公路修建问题

一道假的蓝题，真的假

讲道理我没看题解乱搞的就过了

一看就知道要二分答案啊。

初始化直接对所有的边排下序，贪心地取一号路，用并查集维护一下连通性就能建出生成树。我们要求的其实就是生成树。

方案因为有spj所以乱搞就可以了。我用vector再套pair，排一下第一维就可以了。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<utility>
#include<vector>
const int maxn = 10005, maxm = 20005;
struct edge
{
    int u, v, c1, c2, id;
    bool operator < (const edge &rhs) const
    {
        if(c1 == rhs.c1) return c2 < rhs.c2;
        return c1 < rhs.c1;
    }
} s[maxm];
int father[maxn];
int n, m, k;
std::vector<std::pair<int,int> > answer, temp;
int find(int x)
{
    if(father[x] == x) return x;
    return father[x] = find(father[x]);
}
void merge(int x, int y)
{
    x = find(x); y = find(y);
    if(x != y) father[x] = y;
}
bool check(int mid)
{
    temp.clear();
    for(int i = 1; i <= n; i++) father[i] = i;
    int kk = 0, cnt = 0;
    for(int i = 1; i < m; i++)
    {
        if(find(s[i].u) == find(s[i].v)) continue;
        if(s[i].c1 <= mid)
        {
            kk++; cnt++;
            merge(s[i].u, s[i].v);
            temp.push_back(std::make_pair(s[i].id, 1));
        }
        else if(s[i].c2 <= mid)
        {
            cnt++;
            merge(s[i].u, s[i].v);
            temp.push_back(std::make_pair(s[i].id, 2));
        }
    }
    if(cnt == n - 1 && kk >= k) return true;
    return false;
}
int main()
{
    scanf("%d%d%d", &n, &k, &m);
    int left = 30005, right = -1, ans = -1;
    for(int i = 1; i < m; i++)
    {
        scanf("%d%d%d%d", &s[i].u, &s[i].v, &s[i].c1, &s[i].c2);
        s[i].id = i;
        left = std::min(left, s[i].c2);
        right = std::max(right, s[i].c1);
    }
    std::sort(s + 1, s + m);
    while(left <= right)
    {
        int mid = (left + right) >> 1;
        if(check(mid))
        {
            ans = mid, right = mid - 1;
            answer = temp;
        }
        else left = mid + 1;
    }
    printf("%d\n", ans);
    // ni de fang an
    std::sort(answer.begin(), answer.end());
    for(std::vector<std::pair<int,int> >::iterator it = answer.begin(); it != answer.end(); ++it)
    {
        printf("%d %d\n", it->first, it->second);
    }
    return 0;
}