LA4043 - Ants（二分图完备最佳匹配KM）

Posted 2020-09-28 lytwajue

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option=com_onlinejudge&Itemid=8&page=show_problem&problem=2044">https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2044

大致题意：

平面上有n个白点和n个黑点，求一种完美匹配使他们间的连线不相交

思路：要注意到，若有两种匹配相交，总能够当成对角线补成四边形，然后选四边形的两个边作为匹配就不会相交，并且一定匹配后的距离和缩短了，简单的几何知识，显然最小权匹配不会出现相交情况

防止被卡精度用了龙龙

//	Accepted	C++11	0.079
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
#define rep(i,n) for ( int i=0; i<int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<ll,ll> pii;

template <class T>
inline bool RD(T &ret) {
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != '-' && (c<'0' || c>'9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
template <class T>
inline void PT(T x) {
    if (x < 0) {
        putchar('-');

        x = -x;
    }
    if (x > 9) pt(x / 10);
    putchar(x % 10 + '0');
}


const int N = 100+10;
const ll inf = 1LL<<50;
pii poix[N],poiy[N];
int n;
ll mp[N][N];

ll dis(int x,int y){
        ll a = poix[x].X-poiy[y].X, b = poix[x].Y-poiy[y].Y;
        return (a*a+b*b);
}

int link[N];
ll lx[N],ly[N]; //y中各点匹配状态，x,y中的点标号
ll sla[N];
bool visx[N],visy[N];
bool DFS(int x)
{
        visx[x] = true;
        REP(y,n){
                if(visy[y])continue;
                ll tmp = lx[x] + ly[y] - mp[x][y];
                if(tmp == 0){
                        visy[y] = true;
                        if(link[y] == -1 || DFS(link[y])){
                                link[y] = x;
                                return true;
                        }
                }
                else if(sla[y] > tmp) sla[y] = tmp;
        }
        return false;
}
ll KM()
{
        memset(link,-1,sizeof(link));
        memset(ly,0,sizeof(ly));
        REP(i,n){
                lx[i] = -inf;
                REP(j,n) lx[i] = max(lx[i],mp[i][j]);
        }
        REP(x,n){
                REP(i,n) sla[i] = inf;
                while(true)
                {
                        memset(visx,false,sizeof(visx));
                        memset(visy,false,sizeof(visy));
                        if(DFS(x)) break;
                        ll d = inf;
                        REP(i,n) if(!visy[i]) d = min(d,sla[i]);
                        REP(i,n){
                                if(visx[i]) lx[i] -= d;
                                if(visy[i])ly[i] += d;
                                else sla[i] -= d;
                        }
                }
        }
        ll res = 0;
        REP(i,n) if(link[i] != -1) res += mp[link[i]][i];
        return res;
}

int main(){
        bool flag = 0;
        while(~scanf("%d",&n)){
                if(flag) puts("");
                flag = 1;
                REP(i,n) RD(poix[i].X),RD(poix[i].Y);
                REP(i,n) RD(poiy[i].X),RD(poiy[i].Y);
                REP(x,n) REP(y,n) mp[y][x] = -sqrt(dis(x,y))*10000000;
                KM();
                REP(i,n) printf("%d\n",link[i]);
        }
        //Print a blank line between datasets.
}