bzoj 3158 千钧一发

Description

Input

第一行一个正整数N。

第二行共包括N个正整数，第 个正整数表示Ai。

第三行共包括N个正整数，第 个正整数表示Bi。

Output

共一行，包括一个正整数，表示在合法的选择条件下，可以获得的能量值总和的最大值。

Sample Input

4

3 4 5 12

9 8 30 9

Sample Output

39

HINT

1<=N<=1000,1<=Ai,Bi<=10^6

题解

和bzoj3275几乎一样。。流量是b[i]

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define N 2010
using namespace std;
const int inf = 0x3f3f3f3f;
struct flows {
    struct node {
        int to, next, flow;
    } e[N * N];
    int tot, st[N], dis[N], cur[N];
    void init() {
        tot = -1, memset(e, -1, sizeof e),
        memset(st, -1, sizeof st);
    }
    void add(int x, int y, int z) {
        e[++tot].to = y;
        e[tot].flow = z;
        e[tot].next = st[x];
        st[x] = tot;
    }
    void add_edge(int x, int y, int z) {
//      printf("add%d %d %d\n", x, y, z);
        add(x, y, z), add(y, x, 0);
    }
    queue <int> que;
    int bfs(int S, int T) {
        memcpy(cur, st, sizeof st);
        memset(dis, 0, sizeof dis);
        while (!que.empty()) que.pop();
        que.push(S);
        dis[S] = 1;
        while(!que.empty()) {
            int now = que.front();
            que.pop();
            for (int i = st[now]; ~i; i = e[i].next)
                if (e[i].flow && !dis[e[i].to]) {
                    dis[e[i].to] = dis[now] + 1;
                    if (e[i].to == T) return 1;
                    que.push(e[i].to);
                }
        }
        return 0;
    }
    int finds(int now, int T, int flow) {
        if (now == T)
            return flow;
        int f;
        for (int i = cur[now]; ~i; i = e[i].next) {
            cur[now] = i;
            if (e[i].flow && dis[e[i].to] == dis[now] + 1 &&
                (f = finds(e[i].to, T, min(flow, e[i].flow)))) {
                e[i].flow -= f;
                e[i^1].flow += f;
                return f;
            }
        }
        return 0;
    }
    int dinic(int S, int T) {
        int flow = 0, x;
        while(bfs(S, T)) {
            while(x = finds(S, T, inf)) {
                flow += x;
            }
        }
        return flow;
    }
}flow;
int n, m, a[N], b[N], ans;
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
main() {
    scanf("%d", &n);
    flow.init();
    int S = 0, T = n + 1;
    for (int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &b[i]), ans += b[i];
        if (a[i] & 1)
            flow.add_edge(S, i, b[i]);
        else
            flow.add_edge(i, T, b[i]);
    }
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j) {
            long long t = (long long)a[i] * a[i] + (long long)a[j] * a[j], s = (int)sqrt(t);
            if (s * s == t && gcd(a[i], a[j]) == 1)
                if (a[i] & 1)
                    flow.add_edge(i, j, inf);
                else
                    flow.add_edge(j, i, inf);
        }
    printf("%d\n", ans - flow.dinic(S, T));
}