luogu P2619 [国家集训队2]Tree I

题目链接

luogu P2619 [国家集训队2]Tree I

题解

普通思路就不说了二分增量，生成树check
说一下坑点
二分时，若黑白边权有相同，因为权值相同优先选白边，若在最有增量时出现黑白等权边则更新出 > 和 = 最小值等价，那么不会更新到 = 情况，
因为等价，那么处理时只需看做把等价的黑白两边交换即可
需要每次直接减去 增量 * need 的价值

代码

#include<cstdio>
#include<algorithm> 
const int maxn = 100007; 

int E,V,need;
struct node { 
    int u,v,w,col;
    bool operator < (const node & a) const {
        if(a.w == w) return col < a.col;
        else return w < a.w; 
    } 
}edge[maxn],e[maxn]; 
int ans = 0; 
int fa[maxn]; 
int find(int x) { 
    if(fa[x] != x) fa[x] = find(fa[x]); 
    return fa[x];  
} 
bool check(int x) { 
    int tmp = 0; 
    for(int i = 1;i <= V;++ i) { 
        e[i] = edge[i]; 
    if(e[i].col == 0) e[i].w = edge[i].w + x; 
    }  
    for(int i = 0;i <= E + 1;++ i) fa[i] = i; 
    std::sort(e + 1,e + V + 1);int bs = 0; 
    for(int num = 0,i = 1;i <= V;++ i) { 
        int X = e[i].u,y = e[i].v; 
        int fx = find(X),fy = find(y); 
        if(fx != fy) { 
            tmp += e[i].w;  
            num ++;  
            fa[fx] = fy; 
            if(e[i].col == 0) bs ++; 
        } 
        if(num == E - 1) break;  
    } 
    if(bs >= need) { 
        ans = tmp - (x * need);   
        return true; 
    } 
    return false; 
} 
int main() { 
    scanf("%d%d%d",&E,&V,&need); 
    for(int i = 1;i <= V;++ i) { 
        scanf("%d%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w,&edge[i].col); 
    } 	
    int l = -107,r = 107; 
    while(l <= r) { 
        int mid = (l + r) >> 1; 
        if(check(mid)) l = mid + 1; 
        else r = mid - 1; 
    } 
    printf("%d
",ans); 
    return 0; 
}