心得
听说南京现场赛出了个KM然后卡kuangbin&&lrj板子...
吓得我交一发hdu6346整理个板子，KM的bfs版本
kuangbin的版本直接求是求最大权匹配，
而该板子直接求是求最小权匹配
第一个的是最好用的，不过还是多带几个板子，万一被卡了呢……
代码1
//KM模板
//本题hdu6346 用于求二分图最大权匹配
//如果边权不取反 则用于求二分图最小权匹配
//否则边权取反之后再对答案取反 用于求二分图最大权匹配
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=205;
const ll INF=0x3f3f3f3f3f3f3f3fll;
int t,n;
struct KM{
int n,m[N],way[N];
ll w[N][N],lx[N],ly[N],sl[N];
bool u[N];
void init(){
scanf("%d",&n);
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j){
scanf("%lld",&w[i][j]);
w[i][j]=-w[i][j];
}
}
}
void hungary(int x){
m[0]=x;
int j0=0;
fill(sl,sl+n+1,INF);
fill(u,u+n+1,0);
do{
u[j0]=1;
int i0=m[j0],j1=0;
ll d=INF;
for(int j=1;j<=n;++j)
if(u[j]==0){
ll cur=-w[i0][j]-lx[i0]-ly[j];
if(cur<sl[j]){sl[j]=cur;way[j]=j0;}
if(sl[j]<d){d=sl[j];j1=j;}
}
for(int j=0;j<=n;++j){
if(u[j]){lx[m[j]]+=d;ly[j]-=d;}
else sl[j]-=d;
}
j0=j1;
}while(m[j0]!=0);
do{
int j1=way[j0];m[j0]=m[j1];j0=j1;
}while(j0);
}
ll solve(){
for(int i=1;i<=n;++i)m[i]=lx[i]=ly[i]=way[i]=0;
for(int i=1;i<=n;++i)hungary(i);
ll sum=0;
for(int i=1;i<=n;++i)sum+=w[m[i]][i];
return sum;
}
}q;
int main()
{
scanf("%d",&t);
for(int c=1;c<=t;++c){
q.init();
printf("Case #%d: %lld\n",c,-q.solve());
}
return 0;
}
代码2
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define inf 1000000000000ll
inline int read() {
char ch = getchar(); int x = 0, f = 1;
while(ch < '0' || ch > '9') {
if(ch == '-') f = -1;
ch = getchar();
}
while('0' <= ch && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
namespace KM{
#define M 410
int n;
LL A[M], B[M], mn[M];
int w[M][M], lk[M], way[M];
bool vis[M];
inline void km(){
memset(lk, -1, sizeof(lk));
for(int x = 1; x <= n; ++ x) {
lk[0] = x;
int j0 = 0;
memset(vis, 0, sizeof(vis));
memset(mn, 0x3f, sizeof(mn));
do{
vis[j0] = true;
int i0 = lk[j0], j1;
LL num = inf;
for(int j = 1; j <= n; ++ j){
if(!vis[j]){
LL t = A[i0] + B[j] - w[i0][j];
if(t < mn[j]) mn[j] = t, way[j] = j0;
if(mn[j] < num) num = mn[j], j1 = j;
}
}
for(int j = 0; j <= n; ++ j){
if(vis[j]) A[lk[j]] -= num, B[j] += num;
else mn[j] -= num;
}
//cerr << x <<endl;
j0 = j1;
} while(~lk[j0]);
do{
int j1 = way[j0];
lk[j0] = lk[j1];
j0 = j1;
} while(j0);
}
}
inline LL getsum(){
LL ret = 0;
for(int i = 1; i <= n; ++ i) ret += A[i], ret += B[i];
return ret;
}
}
signed main() {
//int TT = clock();
int T;
scanf("%d", &T);
for(int Ca = 1; Ca <= T; ++ Ca) {
scanf("%d", &KM::n);
for(int i = 1; i <= KM::n; ++ i) {
for(int j = 1; j <= KM::n; ++ j) {
KM::w[i][j] = read();
KM::w[i][j] *= -1;
}
}
//cerr << 2333 << endl;
KM::km();
printf("Case #%d: %I64d\n", Ca, -KM::getsum());
}
//cerr << clock() - TT;
}
代码3
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=207,NPOS=-1;
const ll INF=0x3f3f3f3f3f3f3f3f;
struct Matrix
{
int n;
ll a[N][N];
};
struct KuhnMunkres:Matrix
{
ll hl[N],hr[N],slk[N];
int fl[N],fr[N],vl[N],vr[N],pre[N],q[N],ql,qr;
int check(int i)
{
if(vl[i]=1,fl[i]!=NPOS)return vr[q[qr++]=fl[i]]=1;
while(i!=NPOS)swap(i,fr[fl[i]=pre[i]]);
return 0;
}
void bfs(int s)
{
fill(slk,slk+n,INF);
fill(vl,vl+n,0);
fill(vr,vr+n,0);
q[ql=0]=s;
vr[s]=qr=1;
for(ll d;;)
{
for(; ql<qr; ++ql)
for(int i=0,j=q[ql]; i<n; ++i)
if(d=hl[i]+hr[j]-a[i][j],!vl[i]&&slk[i]>=d)
if(pre[i]=j,d)slk[i]=d;
else if(!check(i))return;
d=INF;
for(int i=0; i<n; ++i)
if(!vl[i]&&d>slk[i])d=slk[i];
for(int i=0; i<n; ++i)
{
if(vl[i])hl[i]+=d;
else slk[i]-=d;
if(vr[i])hr[i]-=d;
}
for(int i=0; i<n; ++i)
if(!vl[i]&&!slk[i]&&!check(i))return;
}
}
void ask()
{
fill(pre,pre+n,NPOS);
fill(fl,fl+n,NPOS);
fill(fr,fr+n,NPOS);
fill(hr,hr+n,0);
for(int i=0; i<n; ++i)hl[i]=*max_element(a[i],a[i]+n);
for(int j=0; j<n; ++j)bfs(j);
}
} km;
int main()
{
int n,nl,nr;
int QAQ,kase=0;
scanf("%d",&QAQ);
while(QAQ--)
{
scanf("%d",&n);
nl=n,nr=n;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%lld",&km.a[i][j]);
km.a[i][j]=-km.a[i][j];
}
}
km.n=max(nl,nr);
km.ask();
long long ans=0;
for(int i=0; i<nl; ++i)
if(km.fl[i]!=NPOS)
ans+=km.a[i][km.fl[i]];
printf("Case #%d: %lld\n",++kase,-ans);
}
}