题目大意就是给定n个slide和n个数的坐标,问是否能唯一确定每个slide又个唯一的数字,数字只能用一次。
首先一个数字能代表一个slide的前提就是这个数字在slide内,预处理下就好了。
然后就是先跑遍二分图,进行一次匹配,在看每个匹配是不是一定只能这样,也就是说slide A是不是只能与数字B匹配。。。
判定的条件就是先去掉匹配关系与可匹配关系(图),然后看是不是这种更改后slide A仍然能够与其他数匹配(增广路),不能就说明是ok的。。。
/***************************************** Author :Crazy_AC(JamesQi) Time :2015 File Name : *****************************************/ // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <iomanip> #include <sstream> #include <string> #include <stack> #include <queue> #include <deque> #include <vector> #include <map> #include <set> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <climits> using namespace std; #define MEM(x,y) memset(x, y,sizeof x) #define pk push_back typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> ii; typedef pair<ii,int> iii; const double eps = 1e-10; const int inf = 1 << 30; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const int N = 30; int g[N][N]; struct point{ int x, y; }p[N]; struct slide{ int xmin,xmax,ymin,ymax; bool contain(const point& rhs) { return xmin < rhs.x && xmax > rhs.x && ymin < rhs.y && ymax > rhs.y; } }s[N]; int n; bool vis[N]; int linker[N]; bool dfs(int u) { for (int i = 1;i <= n;++i) { if (!vis[i] && g[u][i]) { vis[i] = true; if (linker[i] == -1 || dfs(linker[i])) { linker[i] = u; return true; } } } return false; } int solve() { memset(linker, -1,sizeof linker); int ans = 0; for (int i = 1;i <= n;++i) { memset(vis, false,sizeof vis); if (dfs(i)) ans++; } return ans; } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int icase = 0; while(scanf("%d",&n) && n) { for (int i = 1;i <= n;++i) scanf("%d%d%d%d",&s[i].xmin,&s[i].xmax,&s[i].ymin,&s[i].ymax); for (int i = 1;i <= n;++i) scanf("%d%d",&p[i].x,&p[i].y); memset(g, 0,sizeof g); for (int i = 1;i <= n;++i) { for (int j = 1;j <= n;++j) if (s[i].contain(p[j])) g[j][i] = 1; } solve(); bool first = true; if (icase) puts(""); printf("Heap %d\n", ++icase); // for (int i = 1;i <= n;++i) // printf("%4d",linker[i]); // puts(""); for (int i = 1;i <= n;++i) { int x = linker[i]; g[x][i] = 0; linker[i] = -1; memset(vis, false,sizeof vis); if (!dfs(x)) { if (!first) putchar(' '); if (first) first = false; printf("(%c,%d)", i + 'A' - 1, x); linker[i] = x; } g[x][i] = 1; // linker[i] = x; } if (!first) puts(""); if (first) printf("none\n"); //puts("");//puts(""); } return 0; }