BZOJ 3158: 千钧一发(最小割)

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3158: 千钧一发

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 648 Solved: 249
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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

HINT

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

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <algorithm>#define LL long long using namespace std;const int MAXN = 2000000 + 10;const int INF = 0x7fffffff;inline int read(){    int x = 0, f = 1; char ch = getchar();    while(ch < '0' || ch > '9'){if(ch == '-') f=-1; ch = getchar();}    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0'; ch = getchar();}    return x * f;}struct Edge{int to, next, cap, flow;}edge[MAXN<<2];int tot, head[MAXN];int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];void init(){tot = 0; memset(head, -1, sizeof(head));}void addedge(int u, int v, int w, int rw = 0){    edge[tot].to = v; edge[tot].cap = w; edge[tot].next = head[u];    edge[tot].flow = 0; head[u] = tot++;    edge[tot].to = u; edge[tot].cap = rw; edge[tot].next = head[v];    edge[tot].flow = 0; head[v] = tot++;}long long sap(int start, int end, int N){    memset(gap, 0, sizeof(gap));    memset(dep, 0, sizeof(dep));    memcpy(cur, head, sizeof(head));    int u = start; pre[u] = -1; gap[0] = N; long long ans = 0;    while(dep[start] < N)    {        if(u == end)        {            long long Min = INF;            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])                if(Min > edge[i].cap - edge[i].flow)                    Min = edge[i].cap - edge[i].flow;            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])            {                edge[i].flow += Min;                edge[i^1].flow -= Min;            }            u = start; ans += Min; continue;        }        bool flag = false; int v;        for(int i=cur[u];i!=-1;i=edge[i].next)        {            v = edge[i].to;            if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])            {                flag = true;                cur[u] = pre[v] = i;                break;            }        }        if(flag){u = v; continue;}        int Min = N;        for(int i=head[u];i!=-1;i=edge[i].next)            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)            {                Min = dep[edge[i].to];                cur[u] = i;            }        gap[dep[u]]--;        if(!gap[dep[u]]) return ans;        dep[u] = Min + 1;        gap[dep[u]]++;        if(u != start) u = edge[pre[u]^1].to;    }    return ans;}int n, a[1010], b[1010];bool judge(int x, int y){    x = a[x], y = a[y];    if(__gcd(x, y) > 1) return 1;    long long tmp = 1ll * x * x + 1ll * y * y;    long long t = sqrt(tmp);    if(t * t != tmp) return 1;    return 0;}int main(){    n = read();long long ans = 0;init();    int S = 0, T = n + 1;    for(int i=1;i<=n;i++) a[i] = read();    for(int i=1;i<=n;i++) b[i] = read(), ans += b[i];    for(int i=1;i<=n;i++)    {        if(a[i] % 2 == 0) addedge(S, i, b[i]);        else addedge(i, T, b[i]);        for(int j=1;j<=n;j++)        {            if(a[i] % 2 == 0 && a[j] % 2 == 1)            {                if(!judge(i, j)) addedge(i, j, INF);            }        }    }    long long res = sap(S, T, n + 2);    printf("%lld\n", ans - res);    return 0;}

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