UVALive 6525 Attacking rooks 二分匹配经典题

全栈程序员-站长 • 2022年1月31日下午9:00 • 未分类 • 阅读 49

大家好，又见面了，我是全栈君。

题目链接：

<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[N], 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[N], pn;
int tim, T[N];

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。假设能 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;
}
/*
5
X....
X....
..X..
.X...
....X

3
.X.
XXX
XXX

3
.X.
X.X
XXX

3
.X.
X.X
X.X

3
.X.
X.X
.X.
3
XXX
XXX
XXX
15
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX

*/

发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/115695.html原文链接：https://javaforall.net