POJ 1486 Sorting Slides (二分图关键匹配边)

题意

给你n个幻灯片，每个幻灯片有个数字编号1~n，现在给每个幻灯片用A~Z进行编号，在该幻灯片范围内的数字都可能是该幻灯片的数字编号。问有多少个幻灯片的数字和字母确定的。

思路

确定幻灯片的数字就是求完美匹配也就是最大匹配，而题目要求的边就是匹配的关键边，也叫必须边，即任意一个最大匹配一定要包含这条边。

关键边求法：先求一遍最大匹配，然后枚举删去匹配边，看之后的最大匹配是否减小，如果减小则该边是匹配关键边。

这让我想起了寻找关键割边：如果一个割边增加流量后整个最大流增加则该割边是关键割边。呵呵，是不是很像？算法的魅力之一就在于举一反三吧^_^~

代码

#include
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 55;                   //N1+N2
vector  adj[MAXV];
struct MaximumMatchingOfBipartiteGraph{
int vn;
void init(int n){                   //二分图两点集点的个数
vn = n;
for (int i = 0; i <= vn; i ++)     adj[i].clear();
}
void add_uedge(int u, int v){
adj[u].push_back(v);
adj[v].push_back(u);
}
bool vis[MAXV];
int mat[MAXV];                      //记录已匹配点的对应点
bool cross_path(int u){
for (int i = 0; i < (int)adj[u].size(); i ++){
int v = adj[u][i];
if (!vis[v]){
vis[v] = true;
if (mat[v] == 0 || cross_path(mat[v])){
mat[v] = u;
mat[u] = v;
return true;
}
}
}
return false;
}
int hungary(){
mem(mat, 0);
int match_num = 0;
for (int i = 1; i <= vn; i ++){
mem(vis, 0);
if (!mat[i] && cross_path(i)){
match_num ++;
}
}
return match_num;
}
}match;
struct xy{
int x1, x2, y1, y2;
int x, y;
}a[MAXV];
bool del[MAXV][MAXV];
void build(int n){
match.init(n+n);
for (int i = 1; i <= n; i ++){
for (int j = n+1; j <= n+n; j ++){
if (!del[i][j] && a[j].x >= a[i].x1 && a[j].x <= a[i].x2 && a[j].y >= a[i].y1 && a[j].y <= a[i].y2){
match.add_uedge(i, j);
}
}
}
}
map  key_match;
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n;
int t = 1;
while(scanf("%d", &n), n){
for (int i = 1; i <= n; i ++){
scanf("%d %d %d %d", &a[i].x1, &a[i].x2, &a[i].y1, &a[i].y2);
}
for (int i = 1; i <= n; i ++){
scanf("%d %d", &a[i+n].x, &a[i+n].y);
}
mem(del, false);
build(n);
int max_match = match.hungary();
key_match.clear();
for (int i = 1; i <= n; i ++){
key_match.insert(make_pair(i+64, match.mat[i]-n));
}
map  :: iterator it;
for (it = key_match.begin(); it != key_match.end(); it ++){
del[it->first - 64][it->second+n] = true;
build(n);
del[it->first - 64][it->second+n] = false;
int tmp_match = match.hungary();
if (tmp_match == max_match){
it->second = -1;
}

}
printf("Heap %d\n", t ++);
bool ok = 0;
for (it = key_match.begin(); it != key_match.end(); it ++){
if (it->second == -1)   continue;
ok = 1;
printf("(%c,%d) ", it->first, it->second);
}
if (ok){
puts("\n");
}
else{
puts("none\n");
}
}

return 0;
}