晴朗的博客

#include#includeint link[600],mark[600],map[600][600],m,n;int find(int u) {int i;for(i=1;iif(map[u][i]==1&&mark[i]==0) {mark[i]=1;if(link[i]==0||find(link[i])) {link[i]=u;return 1;

1001 Easy Finding POJ-37401002 Power Stations HDOJ-36631003 Treasure Map ZOJ-32091004 Lamp HDOJ-28281005 whosyourdaddy HDOJ-34981006 Bomberman - Just Search! HDOJ-35291007 Squa

/**/#include#include#include#includeusing namespace std;#define inf 0x3fffffff#define N 500struct node{ int u,v,w,f,next;} bian[N*N*10];char ss[N][N];int ma[N][N];int manha(int i,int

转载地址：http://www.cnblogs.com/Griselda/archive/2012/09/17/2689345.html题意：给出每个人适合住的星球信息和该星球能住多少人 第一行给出n m 代表有 n 个人 m 个星球然后接下来n行每行m个数字 1代表适合第 i 个星球 0 代表不适合第 i 个星球最后一行m个数表示第 i 个星球最多可以住多少个人问是不是所有

#include#include#include#define N  400#define inf 999999999int Max(int a,int b ) {return a>b?a:b;}int Min(int a,int b) {return a>b?b:a;}int map[N][N],lx[N],ly[N],s[N],t[N],link[N

#include#include#include#define N 200#define inf 999999999int Max(int a,int b ) {    return a>b?a:b;}int Min(int a,int b) {    return a>b?b:a;}int map[N][N],lx[N],ly[N],s[N],t[N]

【HDU】2255 奔小康赚大钱1533 Going Home1853 Cyclic Tour3488 Tour3435 A new Graph Game2426 Interesting Housing Problem2853 Assignment3718 Similarity3722 Card Game3

#include#include#define N  400#define inf 0x7fffffffint Max(int a,int b ) {return a>b?a:b;}int Min(int a,int b) {return a>b?b:a;}int map[N][N],lx[N],ly[N],s[N],t[N],link[N],n;int

通俗匹配，多重匹配 【HDU】 1068 Girls and Boys1150 Machine Schedule1151 Air Raid1179 Ollivanders: Makers of Fine Wands since 382 BC.1281 棋般游戏14

/*1A 32ms*/#include#include#define N 300int n;struct node {int u,v,next;}bian[N*N*2];int color[N],vis[N],link[N],visit[N],ma[N][N],f[N],head[N],yong;void addedge(int u,int v) {bian[yong].u=u

#include#include#define N 300#define inf 0x3fffffffint mark[N],link[N],n,m;int map[N][N];int find(int u) {int i;for(i=1;i    if(!mark[i]&&map[u][i]) {        mark[i]=1;        if

#include#include//这个分开后男的站在一边女的站在一边，不肯能有les或者gay。最大独立集=n-最大匹配数#define  N  510int map[N][N],n,mark[N],link[N];int find(int u) { int i; for(i=0;i if(!mark[i]&&map[u][i]) {    mark[i]=1;

#include#include#define  N  510int map[N][N],n,mark[N],link[N];int find(int u) { int i; for(i=1;i if(!mark[i]&&map[u][i]) {    mark[i]=1;    if(link[i]==-1||find(link[i])) {       

#include#include#define N 200int map[N][N],visit[N],link[N],n,m;int find(int u) {int i;for(i=1;iif(map[u][i]&&visit[i]==0) {visit[i]=1;if(link[i]==0||find(link[i])) {link[i]=u;re

/*题意：给你一些冰岛。公共海域和浮冰，冰岛可以隔开两个公共海域，浮冰无影响求选尽可能多的选一些公共海域点每行每列仅能选一个。限制条件：冰山可以隔开这个限制条件。即*#*可以选两个预处理：*******#*#***** 可以按行转化 *******#ooooo*#*****按列转化***0**o**ooooooooo*oo**o**o*因为每行每列顶多可