[Tyvj] 1982 weapons allocation (cost flow)

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Question at a glance ..

The connection capacity of each source is 1 and the cost is 0.

Each person connects to a transit node with a capacity of 1 at a cost of 0 (you can also directly connect to the back end without connection)

The transit node Na wants to connect all items A with a capacity of 1 at a cost of 0.

All items A are connected to all items B. The cost is (m-N) ^ 2.

All items B are linked to the side with a capacity of 1 and a cost of 0.

Run a billing flow.

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=1010, M=2000000, oo=1000000000;int ihead[N], cnt=1, d[N], p[N], n, a, b, wa[N], wb[N], vis[N], q[N], front, tail;struct ED { int from, to, cap, w, next; } e[M];inline void add(const int &u, const int &v, const int &c, const int &w) {e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u; e[cnt].cap=c; e[cnt].w=w;e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v; e[cnt].cap=0; e[cnt].w=-w;}inline const bool spfa(const int &s, const int &t) {for1(i, 0, t) d[i]=oo, vis[i]=0;vis[s]=1; d[s]=front=tail=0; q[tail++]=s;int u, v, i;while(front!=tail) {u=q[front++]; if(front==N) front=0;for(i=ihead[u]; i; i=e[i].next) if(e[i].cap && d[v=e[i].to]>d[u]+e[i].w) {d[v]=d[u]+e[i].w; p[v]=i;if(!vis[v]) {vis[v]=1, q[tail++]=v;if(tail==N) tail=0;}}vis[u]=0;}return d[t]!=oo;}int mcf(const int &s, const int &t) {int ret=0, f, u;while(spfa(s, t)) {for(f=oo, u=t; u!=s; u=e[p[u]].from) f=min(f, e[p[u]].cap);for(u=t; u!=s; u=e[p[u]].from) e[p[u]].cap-=f, e[p[u]^1].cap+=f;ret+=d[t]*f;}return ret;}int main() {read(n); read(a); read(b);int s=0, na=1000, t=na+1;for1(i, 1, n) add(s, i, 1, 0), add(i, na, 1, 0);for1(i, 1, a) read(wa[i]);for1(i, 1, b) read(wb[i]);for1(i, 1, a) add(na, n+i, 1, 0);for1(i, 1, b) add(n+a+i, t, 1, 0);for1(i, 1, a) for1(j, 1, b) add(n+i, n+a+j, 1, (wa[i]-wb[j])*(wa[i]-wb[j]));print(mcf(s, t));return 0;}

 

 

 

 

 

Description a batch of weapons (machine guns and armor) delivered by the Logistics force ). You need to assign these weapons to their marine (each of them has a machine gun and a set of armor ). But the problem is...
The models of these weapons are different (the weapons are manufactured by the contractor with the lowest bid ), assign an M-type machine gun and an n-type armor to a Marine. The unsatisfactory value is (m-N) ^ 2 (each marine certainly wants the same type of weapon ).
Your task is to assign a machine gun and B sets of armor to N Marines. Minimum the sum of their dissatisfaction values.
Input Format: inputformat First line: 3 positive integers n, a, B (1 <= n <= A, B <= 80)
Row 2: Number A indicates the model of each machine gun
Row 3: number B indicates the model of each armor set
0 <= model value <= 10000
Output Format outputformat outputs a number: Minimum unsatisfactory value.
Sample input sampleinput [Copy Data]

Sample 1：2 3 39 10 200 10 11Sample 2：3 4 43 9 7 44 2 5 5

Sample output sampleoutput [Copy Data]

Sample 1：2Sample 2：5

[Tyvj] 1982 weapons allocation (cost flow)