poj1486 Sorting Slides

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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        :2015File 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_backtypedef 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;}

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