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[BZOJ2654]tree
试题描述
输入
输出
输入示例
2 2 1 0 1 1 1 0 1 2 0
输出示例
2
数据规模及约定
见“输入”
题解
记得这道题是 WJMZBMR 神题。。。然而被我用类似“[BZOJ3624][Apio2008]免费道路”这道题的方法 rank 1 捉了。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); }
return x * f;
}
#define maxn 50010
#define maxm 100010
int n, m;
struct Edge {
int u, v, w;
bool tp, has;
bool operator < (const Edge& t) const { return w < t.w; }
} es[maxm];
int fa[maxn];
int findset(int x) { return x == fa[x] ? x : fa[x] = findset(fa[x]); }
void init() {
for(int i = 1; i <= n; i++) fa[i] = i;
return ;
}
int main() {
n = read(); m = read(); int need = read();
for(int i = 1; i <= m; i++) {
int u = read() + 1, v = read() + 1, w = read(), tp = read();
es[i] = (Edge){ u, v, w, tp, 0 };
}
sort(es + 1, es + m + 1);
init();
for(int i = 1; i <= m; i++) if(es[i].tp) {
int u = findset(es[i].u), v = findset(es[i].v);
if(u != v) fa[v] = u;
}
for(int i = 1; i <= m; i++) if(!es[i].tp) {
int u = findset(es[i].u), v = findset(es[i].v);
if(u != v) fa[v] = u, es[i].has = 1;
}
init();
int cnt = 0, ans = 0;
for(int i = 1; i <= m; i++) if(es[i].has)
fa[findset(es[i].v)] = findset(es[i].u), ans += es[i].w, cnt++;
for(int i = 1; i <= m && cnt < need; i++) if(!es[i].tp) {
int u = findset(es[i].u), v = findset(es[i].v);
if(u != v) fa[v] = u, ans += es[i].w, cnt++;
}
for(int i = 1; i <= m; i++) if(es[i].tp) {
int u = findset(es[i].u), v = findset(es[i].v);
if(u != v) fa[v] = u, ans += es[i].w;
}
printf("%d\n", ans);
return 0;
}
然而最后又被 wzj 神犇用我的方法超过了。。。
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原文地址:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/5688918.html