UVALive 6525
题目链接：
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=92286#problem/A
题意：
一个n*n象棋棋盘上，有一些障碍点。
问最多放置多少个“车”，使得相邻车不会互相攻击。已知他们不能越过障碍点攻击。
思路：
二分匹配。
对于每一行或者每一列，有几个连续段就裂成几个点，如果该点合法即为’.’，则在两个裂出来的点之间可以连一条边。
源码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#include <stack>
using namespace std;
#define inf (1000000000)
const int MAXN = 500 + 5;
int head[MAXN*MAXN], cnt;
struct Edge
{
int u, v;
int ne, flow;
Edge(){}
Edge(int _u, int _v, int _flow){u = _u, v = _v, flow = _flow, ne = head[_u];}
}edge[MAXN * MAXN * 2];
void add_edge(int u, int v, int flow)
{
edge[cnt] = Edge(u, v, flow);
head[u] = cnt++;
edge[cnt] = Edge(v, u, 0);
head[v] = cnt++;
}
char str[MAXN][MAXN];
int mark1[MAXN][MAXN], mark2[MAXN][MAXN];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
bool valid(int x, int y, int n)
{
for(int i = 0 ; i < 4 ; i++){
int tx = dx[i] + x;
int ty = dy[i] + y;
if(tx < 1 || tx > n) continue;
if(ty < 1 || ty > n) continue;
if(str[tx][ty] == 'X') return false;
}
return true;
}
void build_graph(int n, int s, int t)
{
memset(head, -1, sizeof(head));
cnt = 0;
int pre1 = 0, pre2 = 0;
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++){
if(mark1[i][j] != -1 && mark2[i][j] != -1) //&& valid(i, j, n))
add_edge(mark1[i][j], mark2[i][j], 1);
// if(mark1[i][j] != pre1 && mark1[i][j] != -1){
// pre1 = mark1[i][j];
// add_edge(s, mark1[i][j], 1);
//// pre1 = mark1[i][j];
// }
// if(mark2[i][j] != pre2 && mark2[i][j] != -1){
// pre2 = mark2[i][j];
// add_edge(mark2[i][j], t, 1);
//// pre2 = mark2[i][j];
// }
}
}
}
int d[MAXN*MAXN], vis[MAXN*MAXN], num[MAXN*MAXN];
queue<int>que;
int BFS(int t)
{
memset(vis, 0, sizeof(vis));
memset(num, 0, sizeof(num));
int ans = 0;
d[t] = 0, vis[t] = 1;
while(!que.empty()) que.pop();
que.push(t);
while(!que.empty()){
int u = que.front(); que.pop();
num[d[u]]++;
ans = max(ans, d[u]);
for(int now = head[u] ; now != -1 ; now = edge[now].ne){
int v = edge[now].v;
if(vis[v] == 0) {d[v] = d[u] + 1, vis[v] = 1, que.push(v);}
}
}
return ans;
}
int cur[MAXN*MAXN], p[MAXN*MAXN];
int Augment(int s, int t)
{
int flow = inf;
int u = t;
while(u != s){
flow = min(flow, edge[p[u]].flow);
u = edge[p[u]].u;
}
u = t;
while(u != s){
edge[p[u]].flow -= flow;
edge[p[u]^1].flow += flow;
u = edge[p[u]].u;
}
return flow;
}
int ISAP(int s, int t)
{
// printf("s = %d, t = %d\n", s, t);
int up = BFS(t);
for(int i = s ; i <= t ; i++)
cur[i] = head[i];
int u = s;
int flow = 0;
while(d[s] < 2 * t + 1){
// printf("u = %d\n", u);
if(u == t){
flow += Augment(s, t);
u = s;
}
int ok = 0;
for(int now = cur[u] ; now != -1 ; now = edge[now].ne){
if(edge[now].flow && d[u] == d[edge[now].v] + 1){
int v = edge[now].v;
// printf("first\nv = %d\n", v);
cur[u] = now;
p[v] = now;
u = v;
ok = 1;
break;
}
}
if(!ok){
// printf("second\n");
int ts = 2 * t;
for(int now = head[u] ; now != -1 ; now = edge[now].ne)
if(edge[now].flow){
int v = edge[now].v;
ts = min(ts, d[v]);
}
if(--num[d[u]] == 0) break;
num[d[u] = ts + 1]++;
cur[u] = head[u];
if(u != s) u = edge[p[u]].u;
}
}
return flow;
}
int main()
{
// freopen("UVA 12668.in", "r", stdin);
// freopen("UVA 12668 data2.out", "w", stdout);
int n;
while(scanf("%d", &n) != EOF){
int sum1 = 0, sum2 = 0;
memset(mark1, 0, sizeof(mark1));
memset(mark2, 0, sizeof(mark2));
int pre = 0;
for(int i = 1 ; i <= n ; i++){
scanf("%s", str[i] + 1);
sum1++;
for(int j = 1 ; j <= n ; j++){
if(str[i][j] == 'X' ){
sum1++;
mark1[i][j] = -1;
}
else if(pre != sum1){
pre = sum1;
mark1[i][j] = sum1;
}
else mark1[i][j] = sum1;
// if(pre != str[i][j]){
// pre = str[i][j];
// mark1[i][j] = sum1++;
// }
// else{
// continue;
// }
}
}
sum2 = sum1;
for(int j = 1 ; j <= n ; j++){
sum2++;
for(int i = 1 ; i <= n ; i++){
if(str[i][j] == 'X' ){
sum2++;
mark2[i][j] = -1;
}
else if(pre != sum2){
pre = sum2;
mark2[i][j] = sum2;
}
else mark2[i][j] = sum2;
}
}
// printf("mark1\n");
// for(int i = 1 ; i <= n ; i++){
// for(int j = 1 ; j <= n ; j++)
// printf("%d ", mark1[i][j]);
// printf("\n");
// }
// printf("mark1\n");
// printf("mark2\n");
// for(int i = 1 ; i <= n ; i++){
// for(int j = 1 ; j <= n ; j++)
// printf("%d ", mark2[i][j]);
// printf("\n");
// }
// printf("mark2\n");
// printf("sum1 = %d, sum2 = %d\n", sum1, sum2);
int source = 0, sink = sum2 + 1;
build_graph(n, source, sink);
for(int i = 1 ; i <= sum1 ; i++)
add_edge(0, i, 1);
for(int j = sum1 + 1 ; j <= sum2 ; j++)
add_edge(j, sum2 + 1, 1);
int ans = ISAP(source, sink);
// int temp = 0;
// if(sum1 > sum2 - sum1){
// for(int i = 1 ; i <= n ; i++){
// int f = 1;
// for(int j = 1 ; j <= n ; j++){
// if(mark1[i][j] == -1) {f = 1; continue;}
// else if(f == 1) f = 0, temp++;
// }
// }
// }
// else{
// for(int i = 1 ; i <= n ; i++){
// int f = 1;
// for(int j = 1 ; j <= n ; j++){
// if(mark2[i][j] == -1) {f = 1; continue;}
// else if(f == 1) f = 0, temp++;
// }
// }
// }
// ans = temp - ans;
// ans = max(sum1, sum2) - ans - n;
printf("%d\n", ans);
}
return 0;
}
/*
4
.XX.
.XX.
XX.X
.XX.
*/
