UVALive 6525 Attacking rooks 二分匹配 经典题

题目链接：点击打开链接题意：给定n*n的棋盘，可以在'.'上摆 象棋中的车（X是墙壁）使得任意两个车都不能互相攻击到问：最多能摆多少个车。思路：二分匹配1、若没有X，那么做法就是 X点集为行，Y点集为列，对于图上的每个点所在的行和列(x,y) 建一条边 x->y2、有了X，那么对于每个点所在的上方能接触到的X必须各不相同，所以给每个X标号，第一个X标记成n+13、这样X点集就是行(1-n) 和 n+1-siz (siz是X的个数)4、对于每个点，上方能接触到的最近的X作为列，右方能接触到的最近的Y作为行，建一条边 X->Y而处理每个点上方能接触到的最近的X就是一个dp，右方也是同样处理。然后跑个二分匹配就好。

<pre name="code" class="cpp">#pragma comment(linker, "/STACK:1024000000,1024000000")#include<bits/stdc++.h>template <class T>inline bool rd(T &ret) {    char c; int sgn;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    ret*=sgn;    return 1;}template <class T>inline void pt(T x) {    if (x <0) {        putchar('-');        x = -x;    }    if(x>9) pt(x/10);    putchar(x%10+'0');}using namespace std;const int N = 10105;struct Edge{    int to, nex;}edge[N*2];int head, edgenum;void init(){memset(head, -1, sizeof head); edgenum = 0;}void add(int u, int v){    Edge E = {v, head[u]};    edge[edgenum] = E;    head[u] = edgenum++;}int lef, pn;int tim, T;bool match(int x){    for(int i=head[x]; ~i; i=edge[i].nex)    {        int v = edge[i].to;        if(T[v] != tim)        {            T[v] = tim;            if(lef[v] == -1 || match( lef[v] ))   //match(lef[v]) : 原本连接v的X集点 lef[v] 能不能和别人连，如果能 则v这个点就空出来和x连            {                lef[v] = x;                return true;            }        }    }    return false;}int solve(){    int ans = 0;    memset(lef, -1, sizeof(lef));    for(int i = 1; i<= pn; i++)//X集匹配，X集点标号从 1-pn 匹配边是G[左点].size()    {        tim++;        if( match( i ) ) ans++;    }    return ans;}int n, siz, s[105][105], l[105][105], mp[105][105];char str[105];void input(){	siz = n;	for(int i = 1; i <= n; i++)	{		scanf("%s", str+1);		for(int j = 1; j <= n; j++){			if(str[j] == 'X')				mp[i][j] = ++siz;			else				mp[i][j] = 0;		}	}}void build(){	for(int i = 1; i <= n; i++)		s[0][i] = i;	for(int i = 1; i <= n; i++)		for(int j = 1; j <= n; j++)			if(mp[i][j])				s[i][j] = mp[i][j];			else				s[i][j] = s[i-1][j];	for(int i = 1; i <= n; i++)		l[i][n+1] = i;	for(int i = n; i; i--)    {		for(int j = 1; j <= n; j++)			if(mp[j][i])				l[j][i] = mp[j][i];			else				l[j][i] = l[j][i+1];    }	init();	pn = siz;	for(int i = 1; i <= n; i++)		for(int j = 1; j <= n; j++)			if(mp[i][j] == 0)				add(l[i][j+1], s[i-1][j]);}int main(){    tim = 1; memset(T, 0, sizeof T);	while(cin>>n){		input();		build();		cout<<solve()<<endl;	}	return 0;}/*5X....X......X...X.......X3.X.XXXXXX3.X.X.XXXX3.X.X.XX.X3.X.X.X.X.3XXXXXXXXX15XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX*/