POJ 1770 Special Experiment(树形DP)
2015-01-14 15:20
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题意:太不好理解了,直接看了别人的题解,就是说n个数字,m个数字,在n个数字选出最多的数字,使得任意两个数字差不与m个数字相同,问最多能选出多少数字
思路:乍一看是个最大独立集,把矛盾边连边建图,可事实证明是会超时的,而其实根据题意,可以保证建出来的图是无环的,也就是森林,那么就可以进行简单树形DP了
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cstdlib>
using namespace std;
const int N = 205;
int n, m;
int a
, b
, dp
[2];
vector<int> g
;
void dfs(int u, int p) {
dp[u][0] = 0;
dp[u][1] = a[u];
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == p) continue;
dfs(v, u);
dp[u][0] += max(dp[v][0], dp[v][1]);
dp[u][1] += dp[v][0];
}
}
int main() {
while (~scanf("%d%d", &n, &m) && n || m) {
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
g[i].clear();
}
for (int i = 1; i <= m; i++)
scanf("%d", &b[i]);
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
for (int k = 1; k <= m; k++) {
if (abs(a[i] - a[j]) == b[k]) {
g[i].push_back(j);
g[j].push_back(i);
}
}
}
}
memset(dp, -1, sizeof(dp));
int ans = 0;
for (int i = 1; i <= n; i++) {
if (dp[i][0] == -1) {
dfs(1, 0);
ans += max(dp[i][0], dp[i][1]);
}
}
printf("%d\n", ans);
}
return 0;
}
思路:乍一看是个最大独立集,把矛盾边连边建图,可事实证明是会超时的,而其实根据题意,可以保证建出来的图是无环的,也就是森林,那么就可以进行简单树形DP了
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cstdlib>
using namespace std;
const int N = 205;
int n, m;
int a
, b
, dp
[2];
vector<int> g
;
void dfs(int u, int p) {
dp[u][0] = 0;
dp[u][1] = a[u];
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == p) continue;
dfs(v, u);
dp[u][0] += max(dp[v][0], dp[v][1]);
dp[u][1] += dp[v][0];
}
}
int main() {
while (~scanf("%d%d", &n, &m) && n || m) {
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
g[i].clear();
}
for (int i = 1; i <= m; i++)
scanf("%d", &b[i]);
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
for (int k = 1; k <= m; k++) {
if (abs(a[i] - a[j]) == b[k]) {
g[i].push_back(j);
g[j].push_back(i);
}
}
}
}
memset(dp, -1, sizeof(dp));
int ans = 0;
for (int i = 1; i <= n; i++) {
if (dp[i][0] == -1) {
dfs(1, 0);
ans += max(dp[i][0], dp[i][1]);
}
}
printf("%d\n", ans);
}
return 0;
}
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