poj1192 最优连通子集

额。是单整点集。一开始没看到。。

还算好做，树形dp。

/**
 * Problem:POJ1192
 * Author:Shun Yao
 * Time:2013.5.20
 * Result:Accepted
 * Memo:DP
 */

#include <cstring>
#include <cstdlib>
#include <cstdio>

long abs(long x) {
	return x < 0 ? -x : x;
}

const long Maxn = 1005;

long n, ans, now, f[Maxn];
char visited[Maxn];

class pnode {
public:
	long x, y, c;
	pnode() {}
	~pnode() {}
} p[Maxn];

class gnode {
public:
	long v;
	gnode *next;
	gnode() {}
	~gnode() {}
	gnode(long V, gnode *ne):v(V), next(ne) {}
} *g[Maxn];

void add(long x, long y) {
	g[x] = new gnode(y, g[x]);
	g[y] = new gnode(x, g[y]);
}

void DP(long x) {
	visited[x] = 1;
	f[x] = p[x].c;
	for (gnode *e = g[x]; e; e = e->next) {
		if (!visited[e->v]) {
			DP(e->v);
			if (f[e->v] > 0)
				f[x] += f[e->v];
		}
	}
	if (now < f[x])
		now = f[x];
}

int main() {
	static long i, j;
	freopen("poj1192.in", "r", stdin);
	freopen("poj1192.out", "w", stdout);
	scanf("%ld", &n);
	for (i = 1; i <= n; ++i) {
		g[i] = 0;
		scanf("%ld%ld%ld", &p[i].x, &p[i].y, &p[i].c);
		for (j = 1; j < i; ++j)
			if (abs(p[i].x - p[j].x) + abs(p[i].y - p[j].y) == 1)
				add(i, j);
	}
	memset(visited, 0, sizeof visited);
	for (i = 1; i <= n; ++i)
		if (!visited[i]) {
			now = 0;
			DP(i);
			ans += now;
		}
	printf("%ld", ans);
	fclose(stdin);
	fclose(stdout);
	return 0;
}