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，能够在'.'上摆 象棋中的车（X是墙壁）使得随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，对于图上的每一个点所在的行和列(x,y) 建一条边 x->y2、有了X，那么对于每一个点所在的上方能接触到的X必须各不同样。所以给每一个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、对于每一个点，上方能接触到的近期的X作为列，右方能接触到的近期的Y作为行，建一条边 X->Y而处理每一个点上方能接触到的近期的X就是一个dp。右方也是相同处理。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。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;elsemp[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];elses[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];elsel[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*/