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【贪心】 TOJ 4121. Muxiaokui's Problem

2015-07-20 20:03 381 查看
对两个连通块做最小生成树,然后中间的边选一条代价最小的就行了。。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1005
#define maxm 1000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct node
{
int u, v;
double c;
node(int u = 0, int v = 0, double c = 0) : u(u), v(v), c(c) {}
}e[maxm];

int flag[maxn];
int f[maxn];
pair<double, double> p[maxn];
double tt[maxm];
int cnt, n;

int find(int u)
{
return f[u] = f[u] == u ? f[u] : find(f[u]);
}

int cmp(node aa, node bb)
{
return aa.c < bb.c;
}

bool merge(int u, int v)
{
int uu = find(u), vv = find(v);
if(uu == vv) return false;
f[uu] = vv;
return true;
}

void work()
{
int kk, d;
double u, v;
scanf("%d%d", &d, &n);
cnt = 0;
for(int i = 0; i < n; i++) {
scanf("%lf%lf%d", &u, &v, &kk);
p[i] = mp(u, v);
flag[i] = kk;
}

int tcnt = 0;
double res = 1e15;
for(int i = 0; i < n; i++)
for(int j = i+1; j < n; j++) {
double t1 = p[i].first - p[j].first;
double t2 = p[i].second - p[j].second;
double t = t1 * t1 + t2 * t2;
t = sqrt(t);
if(flag[i] ^ flag[j]) res = min(res, t);
else e[cnt++] = node(i, j, t);
}
if(fabs(res - 1e15) > eps) tt[tcnt++] = res;
sort(e, e+cnt, cmp);
for(int i = 0; i < n; i++) f[i] = i;
for(int i = 0; i < cnt; i++) {
if(merge(e[i].u, e[i].v)) tt[tcnt++] = e[i].c;
}
double ans = 0;
sort(tt, tt+tcnt);
for(int i = 0; i < tcnt - d; i++) ans += tt[i];
printf("%.2f\n", ans);

}

int main()
{
int _;
scanf("%d", &_);
while(_--) work();

return 0;
}
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