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;

}