poj3565Ants【最小权匹配】

大意：

左图为一个坐标轴上的点  其中黑点代表ants 白点代表apple 问怎样安排ants匹配apple才能使人一两条边不想交

分析：

如左图，我们假设a->d,b->c为一个最佳匹配  交点为e

那么ad+bc = ae+ ed + be + ec = (ae + ec) + (be + ed)  该值大于ac+bd

所以匹配为最佳匹配不成立

因此我们只要求出二分图的最小匹配就是结果了

但是wa了；

wa的代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const long long maxn = 105;
const long long INF = 10000000000;


long long W[maxn][maxn];
long long Lx[maxn], Ly[maxn];
long long Left[maxn];
bool S[maxn], T[maxn];

long long n;

bool match(long long i) {
    S[i] = true;
    for(long long j = 1; j <= n; j++) if (Lx[i]+Ly[j] == W[i][j] && !T[j]){
        T[j] = true;
        if (!Left[j] || match(Left[j])){
            Left[j] = i;
            return true;
        }
    }
    return false;
}

void update() {
    long long a = INF;
    for(long long i = 1; i <= n; i++) if(S[i])
        for(long long j = 1; j <= n; j++) if(!T[j])
            a = min(a, Lx[i]+Ly[j] - W[i][j]);
    for(long long i = 1; i <= n; i++) {
        if(S[i]) Lx[i] -= a;
        if(T[i]) Ly[i] += a;
    }
}

void KM() {
    for(long long i = 1; i <= n; i++) {
        Left[i] = Lx[i] = Ly[i] = 0;
        for(long long j = 1; j <= n; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(long long i = 1; i <= n; i++) {
        for(;;) {
            for(long long j = 1; j <= n; j++) S[j] = T[j] = false;
            if(match(i)) break; else update();
        }
    }
}


struct Node {
    long long x, y;
}ant[maxn], apple[maxn];


long long dist(Node n1, Node n2) {
    return (n1.x - n2.x) * (n1.x - n2.x) + (n1.y - n2.y) * (n1.y - n2.y);
}

int main() {
    while(EOF != scanf("%lld",&n)) {
        for(long long i = 1; i <= n; i++) scanf("%lld %lld",&ant[i].x, &ant[i].y);
        for(long long i = 1; i <= n; i++) scanf("%lld %lld",&apple[i].x, &apple[i].y);
        memset(W, 0, sizeof(W));
        for(long long i = 1; i <= n; i++) {
            for(long long j = 1; j <= n; j++) {
                W[i][j] = -dist(apple[i], ant[j]);
            }
        }
        KM();
        for(long long i = 1; i <= n; i++) {
            printf("%lld
",Left[i]);
        }
    }
    return 0;
}

想了好长时间终于知道哪儿错了

我只是用的距离的平方做的比较

也就是说 a^2 + b ^ 2 > c ^ c + d ^ 2 不能推出 a + b > c + d

比如1^2 + 100^2 > 50 ^ 2 + 51 ^2 但是 1 + 100 < 51 + 52

所以平方比较是错的

ac代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const int maxn = 105;
const double INF = 1000000000;

double W[maxn][maxn];
double Lx[maxn], Ly[maxn];
int Left[maxn];
bool S[maxn], T[maxn];

int n;

bool match(int i) {
    S[i] = true;
    for(int j = 1; j <= n; j++) if ( (abs(Lx[i]+Ly[j] - W[i][j]) < 0.00000001) && !T[j]){
        T[j] = true;
        if (!Left[j] || match(Left[j])){
            Left[j] = i;
            return true;
        }
    }
    return false;
}

void update() {
    double a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
        for(int j = 1; j <= n; j++) if(!T[j])
            a = min(a, Lx[i]+Ly[j] - W[i][j]);
    for(int i = 1; i <= n; i++) {
        if(S[i]) Lx[i] -= a;
        if(T[i]) Ly[i] += a;
    }
}

void KM() {
    for(int i = 1; i <= n; i++) {
        Left[i] = 0;
        Lx[i] = Ly[i] = 0.0;
        for(int j = 1; j <= n; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++) {
        for(;;) {
            for(int j = 1; j <= n; j++) S[j] = T[j] = false;
            if(match(i)) break; else update();
        }
    }
}

struct Node {
    double x, y;
}ant[maxn], apple[maxn];

double dist(Node n1, Node n2) {
    return sqrt((n1.x - n2.x) * (n1.x - n2.x) + (n1.y - n2.y) * (n1.y - n2.y));
}

int main() {
    //freopen("3565.txt","r",stdin);
    while(EOF != scanf("%d",&n)) {
        for(int i = 1; i <= n; i++) scanf("%lf %lf",&ant[i].x, &ant[i].y);
        for(int i = 1; i <= n; i++) scanf("%lf %lf",&apple[i].x, &apple[i].y);
        memset(W, 0, sizeof(W));
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                W[i][j] = -dist(apple[i], ant[j]);
            }
        }
        KM();
        for(int i = 1; i <= n; i++) {
            printf("%d
",Left[i]);
        }
    }
    return 0;
}