BZOJ 3624: [Apio2008]免费道路 [生成树 并查集]

题意：

一张图0,1两种边，构造一个恰有k条0边的生成树

---

优先选择1边构造生成树，看看0边是否小于k

然后保留这些0边，补齐k条，再加1边一定能构成生成树

类似kruskal的证明

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=2e4+5, M=1e5+5;
typedef long long ll;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, k, u, v, c, m0, m1, p;
struct meow{int u, v, c;}a[M], b[M], ans[N];
int fa[N];
int find(int x) {return x==fa[x] ? x : fa[x]=find(fa[x]);}
int flag[N];
int main() {
    freopen("in","r",stdin);
    n=read(); m=read(); k=read();
    for(int i=1; i<=m; i++) {
        u=read(), v=read(), c=read();
        if(c==1) a[++m1]=(meow){u,v,c};
        else b[++m0]=(meow){u,v,c};
    }

    int cnt=0;
    for(int i=1; i<=n; i++) fa[i]=i;
    for(int i=1; i<=m1; i++) {
        u=a[i].u, v=a[i].v;
        int x=find(u), y=find(v);
        if(x==y) continue;
        fa[x]=y;
        if(++cnt == n-1) break;
    }
    for(int i=1; i<=m0; i++) {
        u=b[i].u, v=b[i].v;
        int x=find(u), y=find(v);
        if(x==y) continue;
        fa[x]=y; ans[++p]=b[i]; flag[i]=1;
        if(++cnt == n-1) break;
    }
    if(p > k || cnt < n-1) {puts("no solution"); return 0;}

    for(int i=1; i<=n; i++) fa[i]=i;
    for(int i=1; i<=p; i++) fa[find(ans[i].u)] = find(ans[i].v);
    cnt=p;
    if(cnt<k) for(int i=1; i<=m0; i++) if(!flag[i]){
        u=b[i].u, v=b[i].v;
        int x=find(u), y=find(v);
        if(x==y) continue;
        fa[x]=y; ans[++cnt]=b[i];
        if(cnt == k) break;
    }
    for(int i=1; i<=m1; i++) {
        u=a[i].u, v=a[i].v;
        int x=find(u), y=find(v);
        if(x==y) continue;
        fa[x]=y; ans[++cnt]=a[i];
        if(cnt == n-1) break;
    }
    for(int i=1; i<=cnt; i++) printf("%d %d %d
",ans[i].u, ans[i].v, ans[i].c);
}

2017-10-03 今天又写了一下 以前好像有点问题洛谷wa1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1e5+5, M = 1e5+5;
typedef long long ll;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, k;
struct edge {int u, v, c;} e[M];
int flag[N], fa[N], ans[N];
int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}
int main() {
    //freopen("in", "r", stdin);
    scanf("%d %d %d", &n, &m, &k);
    for(int i=1; i<=m; i++) e[i].u = read(), e[i].v = read(), e[i].c = read();
    
    for(int i=1; i<=n; i++) fa[i] = i;
    int num = 0;
    for(int i=1; i<=m; i++) if(e[i].c == 0) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2;
        if(++num == n-1) break;
    }
    int one = 0;
    for(int i=1; i<=m; i++) if(e[i].c == 1) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2;
        one++; ans[++ans[0]] = i; flag[i] = 1;
        if(++num == n-1) break;
    }
    if(one > n-k) {puts("no solution"); return 0;}
    if(num < n-1) {puts("no solution"); return 0;}
     
    for(int i=1; i<=n; i++) fa[i] = i;
    num = 0;
    for(int i=1; i<=m; i++) if(e[i].c == 1) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2;
        if(++num == n-1) break;
    }
    int zero = 0;
    for(int i=1; i<=m; i++) if(e[i].c == 0) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2; 
        zero++; ans[++ans[0]] = i; flag[i] = 1;
        if(++num == n-1) break;
    }
    if(zero > k) {puts("no solution"); return 0;}
    
    for(int i=1; i<=n; i++) fa[i] = i;
    num = ans[0];
    for(int i=1, t; i<=ans[0]; i++) t = ans[i], fa[find(e[t].u)] = find(e[t].v);
    if(zero < k) for(int i=1; i<=m; i++) if(!flag[i] && e[i].c == 0) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2;
        ans[++ans[0]] = i; 
        ++num;
        if(++zero == k) break;
    }
    if(num < n-1) for(int i=1; i<=m; i++) if(!flag[i] && e[i].c == 1) {
        int f1 = find(e[i].u), f2 = find(e[i].v);
        if(f1 == f2) continue;
        fa[f1] = f2;
        ans[++ans[0]] = i;
        if(++num == n-1) break;
    }
    
    if(zero != k || num != n-1) {puts("no solution"); return 0;}
    for(int i=1, t; i<=ans[0]; i++) t = ans[i], printf("%d %d %d
", e[t].u, e[t].v, e[t].c);
}