Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 1329 | Accepted: 668 |
Description
Input
Output
Sample Input
5 0 0 -2 0 1 1 1 0 1 0 -1 1 -1 0 1
Sample Output
2
Source
/*
DP,利用DFS遍历子树
假设f(i)表示以i为结点的子树能够取道的最大值,则
for all i's neighbour j:
res = 0;
f(i) = res + f(j), if f(j) >= 0
最后结果是maxVal = max(f(i)), for all i
这道题有点类似"最大子段问题"
*/
#include <iostream>
#include <vector>
#include <cmath>
#define MAX_N 1000
#define maxv(a, b) ((a) >= (b) ? (a) : (b))
using namespace std;
int maxVals[MAX_N + 1];
bool v[MAX_N + 1];
int maxVal = INT_MIN;
int num;
struct node
{
int x, y, c;
int next[5];
node()
{
x = y = c = 0;
for(int i = 0; i <= 4; i++)
next[i] = 0;
}
}nodes[MAX_N + 1];
bool neighb(const node &p1, const node &p2)
{
return abs(p1.x - p2.x) + abs(p1.y - p2.y) == 1;
}
void init()
{
int i;
for(i = 1; i <= MAX_N; i++)
maxVals[i] = INT_MIN;
}
int dfs(int p)
{
int nNum = nodes[p].next[0], i, res;
int maxV = 0;
for(i = 1; i <= nNum; i++)
{
if(!v[nodes[p].next[i]])
{
v[nodes[p].next[i]] = true;
int curVal = dfs(nodes[p].next[i]);
if(curVal >= 0)
maxV += curVal;
}
}
if(nNum == 0)
maxV = 0;
if(maxV >= 0)
res = maxV + nodes[p].c;
else
res = nodes[p].c;
if(res > maxVal)
maxVal = res;
return res;
}
int main()
{
int i, j;
scanf("%d", &num);
for(i = 1; i <= num; i++)
{
init();
scanf("%d%d%d", &nodes[i].x, &nodes[i].y, &nodes[i].c);
for(j = 1; j < i; j++)
{
if(neighb(nodes[j], nodes[i]))
{
nodes[i].next[0]++;
nodes[i].next[nodes[i].next[0]] = j;
nodes[j].next[0]++;
nodes[j].next[nodes[j].next[0]] = i;
}
}
}
v[1] = true;
dfs(1);
printf("%d/n", maxVal);
return 0;
}