D. Arpa's weak amphitheater and Mehrdad's valuable Hoses 分组背包模板题
http://codeforces.com/problemset/problem/742/D
并查集预处理出所有关系。
一开始的时候,我预处理所有关系后,然后选择全部的时候,另起了一个for,然后再判断。
这样是不对的。因为这样使得同一组里面可能选择了两次。
3 0 2
1 2 3
1 1 3
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL; #include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 2e3 + ;
int w[maxn];
int bea[maxn];
int fa[maxn];
int tofind(int x) {
if (fa[x] == x) return x;
else return fa[x] = tofind(fa[x]);
}
void tomerge(int x, int y) {
x = tofind(x);
y = tofind(y);
fa[y] = x;
}
vector<int>a[maxn];
int dp[ + ];
void work() {
int n, m, tot;
cin >> n >> m >> tot;
for (int i = ; i <= n; ++i) fa[i] = i;
for (int i = ; i <= n; ++i) cin >> w[i];
for (int i = ; i <= n; ++i) cin >> bea[i];
for (int i = ; i <= m; ++i) {
int u, v;
cin >> u >> v;
tomerge(u, v);
}
for (int i = ; i <= n; ++i) {
a[tofind(i)].push_back(i);
}
int tn = n;
for (int i = ; i <= n; ++i) {
if (a[i].size() == ) continue;
int ww = , bb = ;
for (int k = ; k < a[i].size(); ++k) {
ww += w[a[i][k]];
bb += bea[a[i][k]];
}
w[++tn] = ww;
bea[tn] = bb;
a[i].push_back(tn);
}
for (int i = ; i <= n; ++i) {
if (a[i].size() == ) continue;
for (int j = tot; j >= ; --j) {
for (int k = ; k < a[i].size(); ++k) {
if (j >= w[a[i][k]]) {
dp[j] = max(dp[j], dp[j - w[a[i][k]]] + bea[a[i][k]]);
}
}
}
// int ww = 0, bbea = 0;
// for (int k = 0; k < a[i].size(); ++k) {
// ww += w[a[i][k]];
// bbea += bea[a[i][k]];
// }
// for (int j = tot; j >= ww; --j) {
// dp[j] = max(dp[j], dp[j - ww] + bbea);
// }
}
cout << dp[tot] << endl;
} int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return ;
}
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