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：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5
6 const long long maxn = 105;
7 const long long INF = 10000000000;
8
9
10 long long W[maxn][maxn];
11 long long Lx[maxn], Ly[maxn];
12 long long Left[maxn];
13 bool S[maxn], T[maxn];
14
15 long long n;
16
17 bool match(long long i) {
18 S[i] = true;
19 for(long long j = 1; j <= n; j++) if (Lx[i]+Ly[j] == W[i][j] && !T[j]){
20 T[j] = true;
21 if (!Left[j] || match(Left[j])){
22 Left[j] = i;
23 return true;
24 }
25 }
26 return false;
27 }
28
29 void update() {
30 long long a = INF;
31 for(long long i = 1; i <= n; i++) if(S[i])
32 for(long long j = 1; j <= n; j++) if(!T[j])
33 a = min(a, Lx[i]+Ly[j] - W[i][j]);
34 for(long long i = 1; i <= n; i++) {
35 if(S[i]) Lx[i] -= a;
36 if(T[i]) Ly[i] += a;
37 }
38 }
39
40 void KM() {
41 for(long long i = 1; i <= n; i++) {
42 Left[i] = Lx[i] = Ly[i] = 0;
43 for(long long j = 1; j <= n; j++)
44 Lx[i] = max(Lx[i], W[i][j]);
45 }
46 for(long long i = 1; i <= n; i++) {
47 for(;;) {
48 for(long long j = 1; j <= n; j++) S[j] = T[j] = false;
49 if(match(i)) break; else update();
50 }
51 }
52 }
53
54
55 struct Node {
56 long long x, y;
57 }ant[maxn], apple[maxn];
58
59
60 long long dist(Node n1, Node n2) {
61 return (n1.x - n2.x) * (n1.x - n2.x) + (n1.y - n2.y) * (n1.y - n2.y);
62 }
63
64 int main() {
65 while(EOF != scanf("%lld",&n)) {
66 for(long long i = 1; i <= n; i++) scanf("%lld %lld",&ant[i].x, &ant[i].y);
67 for(long long i = 1; i <= n; i++) scanf("%lld %lld",&apple[i].x, &apple[i].y);
68 memset(W, 0, sizeof(W));
69 for(long long i = 1; i <= n; i++) {
70 for(long long j = 1; j <= n; j++) {
71 W[i][j] = -dist(apple[i], ant[j]);
72 }
73 }
74 KM();
75 for(long long i = 1; i <= n; i++) {
76 printf("%lld\n",Left[i]);
77 }
78 }
79 return 0;
80 }
wa的代码
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：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 using namespace std;
6
7 const int maxn = 105;
8 const double INF = 1000000000;
9
10 double W[maxn][maxn];
11 double Lx[maxn], Ly[maxn];
12 int Left[maxn];
13 bool S[maxn], T[maxn];
14
15 int n;
16
17 bool match(int i) {
18 S[i] = true;
19 for(int j = 1; j <= n; j++) if ( (abs(Lx[i]+Ly[j] - W[i][j]) < 0.00000001) && !T[j]){
20 T[j] = true;
21 if (!Left[j] || match(Left[j])){
22 Left[j] = i;
23 return true;
24 }
25 }
26 return false;
27 }
28
29 void update() {
30 double a = INF;
31 for(int i = 1; i <= n; i++) if(S[i])
32 for(int j = 1; j <= n; j++) if(!T[j])
33 a = min(a, Lx[i]+Ly[j] - W[i][j]);
34 for(int i = 1; i <= n; i++) {
35 if(S[i]) Lx[i] -= a;
36 if(T[i]) Ly[i] += a;
37 }
38 }
39
40 void KM() {
41 for(int i = 1; i <= n; i++) {
42 Left[i] = 0;
43 Lx[i] = Ly[i] = 0.0;
44 for(int j = 1; j <= n; j++)
45 Lx[i] = max(Lx[i], W[i][j]);
46 }
47 for(int i = 1; i <= n; i++) {
48 for(;;) {
49 for(int j = 1; j <= n; j++) S[j] = T[j] = false;
50 if(match(i)) break; else update();
51 }
52 }
53 }
54
55 struct Node {
56 double x, y;
57 }ant[maxn], apple[maxn];
58
59 double dist(Node n1, Node n2) {
60 return sqrt((n1.x - n2.x) * (n1.x - n2.x) + (n1.y - n2.y) * (n1.y - n2.y));
61 }
62
63 int main() {
64 //freopen("3565.txt","r",stdin);
65 while(EOF != scanf("%d",&n)) {
66 for(int i = 1; i <= n; i++) scanf("%lf %lf",&ant[i].x, &ant[i].y);
67 for(int i = 1; i <= n; i++) scanf("%lf %lf",&apple[i].x, &apple[i].y);
68 memset(W, 0, sizeof(W));
69 for(int i = 1; i <= n; i++) {
70 for(int j = 1; j <= n; j++) {
71 W[i][j] = -dist(apple[i], ant[j]);
72 }
73 }
74 KM();
75 for(int i = 1; i <= n; i++) {
76 printf("%d\n",Left[i]);
77 }
78 }
79 return 0;
80 }
ac代码
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时间： 2024-09-30 19:50:04