【Codeforces 808F】【网络流】Card Game 题解

F. Card Game
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Digital collectible card games have become very popular recently. So Vova decided to try one of these.

Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can’t place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova’s character.

At the moment Vova’s character’s level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.

Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 100000).

Then n lines follow, each of these lines contains three numbers that represent the corresponding card: pi, ci and li (1 ≤ pi ≤ 1000, 1 ≤ ci ≤ 100000, 1 ≤ li ≤ n).

Output
If Vova won’t be able to build a deck with required power, print  - 1. Otherwise print the minimum level Vova has to reach in order to build a deck.

Examples
input
5 8
5 5 1
1 5 4
4 6 3
1 12 4
3 12 1
output
4
input
3 7
4 4 1
5 8 2
5 3 3
output
2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <stack>
#define INF 2100000000
#define LL long long
#define clr(x) memset(x, 0, sizeof(x))
#define ms(a, x) memset(x, a, sizeof(x))

using namespace std;

template <class T> inline void read(T &x) {
    int flag = 1; x = 0;
    char ch = (char)getchar();
    while(ch <  '0' || ch >  '9') { if(ch == '-')  flag = -1; ch = (char)getchar(); }
    while(ch >= '0' && ch <= '9') { x = (x<<1)+(x<<3)+ch-'0'; ch = (char)getchar(); }
    x *= flag;
}

const int N = 105, M = 50005, NN = 200000;
int n,p[N],C[N],L[N],h[N],e[M],nxt[M],fl[M],m,l,r,mid,ch[M],tot,f[N],visit[N],ans,now[N];
bool bz[NN+5];
int pr[NN+5];
char c;

inline void add(int x, int y, int f, int o) {
    e[++tot] = y; nxt[tot] = h[x]; fl[tot] = f; ch[tot] = tot+o; h[x] = tot;
}

int aug(int x, int F) {
    visit[x] = tot;
    if(x == n+1) return F;
    for(int i = now[x]; i; i = nxt[i])
        if(fl[i] > 0 && f[x] == f[e[i]]+1 && visit[e[i]] < tot) {
            int tmp = aug(e[i], min(F, fl[i]));
            if(tmp > 0) {
                fl[i] -= tmp;
                fl[ch[i]] += tmp;
                now[x] = i;
                return tmp;
            }
        }
    return now[x] = 0;
}

void flow() {
    for(int i = 0; i <= n+1; i++) now[i] = h[i];
    while (1) {
        tot++;
        int tmp = aug(0, INF);
        if(!tmp) return;
        ans -= tmp;
    }
}

bool check() {
    int tmp = INF;
    for(int i = 0; i <= n+1; i++) if (visit[i] == tot)
        for(int j = h[i]; j; j = nxt[j]) if(fl[j] > 0 && visit[e[j]] < tot && f[e[j]]+1-f[i] < tmp)
            tmp = f[e[j]]+1-f[i];
    if(tmp == INF) return 0;
    for(int i = 0; i <= n+1; i++) if(visit[i] == tot) f[i] += tmp;
    return 1;
}

bool Judge(int x) {
    int t = 0; tot = ans = 0; clr(h);
    for(int i = 1; i <= n; i++) if(L[i] <= x && C[i] == 1 && (!t || p[i] > p[t])) t = i;
    for(int i = 1; i <= n; i++) if(L[i] <= x && !(C[i] == 1 && i != t))
        if(C[i]&1) add(0, i, p[i], 1), add(i, 0, 0, -1), ans += p[i];
        else add(i, n+1, p[i], 1), add(n+1, i, 0, -1), ans += p[i];
    for(int i = 1; i <= n; i++) if(L[i] <= x)
        for(int j = 1; j < i; j++) if(L[j] <= x && !bz[C[j]+C[i]])
            if (C[i]&1) add(i, j, INF, 1), add(j, i, 0, -1);
            else add(j, i, INF, 1), add(i, j, 0, -1);
    clr(f); clr(visit); tot = 0;
    for(flow(); check(); flow());
    return ans >= m;
}

int main() {
    for (int i = 2; i <= NN; i++) {
        if(!bz[i]) pr[tot++] = i;
        for(int j = 0; j < tot && i*pr[j] <= NN; j++) {
            bz[i*pr[j]] = 1;
            if(!(i%pr[j])) break;
        }
    }
    read(n); read(m);
    for(int i = 1; i <= n; i++) read(p[i]), read(C[i]), read(L[i]);
    for(l = 1, r = n+1, mid = (l+r)>>1; l < r; mid = (l+r)>>1)
        if (Judge(mid)) r = mid;
        else l = mid+1;
    if(l > n) printf("-1\n");
    else printf("%d\n",l);
    return 0;
}