本文共 4071 字，大约阅读时间需要 13 分钟。
题目描述
输入
输出
输出一个整数a，表示最多能放置炸弹的个数
样例输入
```
4 4
#***
*#**
**#*
xxx#
```
样例输出
5
题解：每一个空位都能引爆一行和列,将行和列拆分出来，进行二分图匹配，当然也可以用网络流来做
二分图匹配：
#include网络流：#include #include #define N 6000#define M 300000using namespace std;int map[60][60] , mp[1000][1000], bx[60][60] , by[60][60] , tx , ty , head[N] , to[M] , val[M] , next[M] , cnt = 1 , s , t , dis[N] , frm[N];char str[60];bool vis[N];bool bfs(int x){ for(int i = 1; i <= ty ; ++i) { if(!mp[x][i] || vis[i]) continue; vis[i] = 1; if(!frm[i] || bfs(frm[i])) { frm[i] = x; return true; } } return false;}int main(){ int n , m , i , j , ans = 0; scanf("%d%d" , &n , &m); for(i = 1 ; i <= n ; i ++ ) { scanf("%s" , str + 1); for(j = 1 ; j <= m ; j ++ ) map[i][j] = (str[j] == '*' ? 0 : str[j] == 'x' ? 1 : 2); } for(i = 1 ; i <= n ; i ++ ) { tx ++ ; for(j = 1 ; j <= m ; j ++ ) { bx[i][j] = tx; if(map[i][j] == 2) tx ++ ; } } for(j = 1 ; j <= m ; j ++ ) { ty ++ ; for(i = 1 ; i <= n ; i ++ ) { by[i][j] = ty; if(map[i][j] == 2) ty ++ ; } } for(i = 1 ; i <= n ; i ++ ) for(j = 1 ; j <= m ; j ++ ) if(!map[i][j]) mp[bx[i][j]][by[i][j]] = 1; for(int i = 1; i <= tx ; ++i) { memset(vis,0,sizeof(vis)); if(bfs(i)) ++ans; } printf("%d\n" , ans); return 0;}
#include#include #include #define N 6000#define M 300000using namespace std;queue q;int map[60][60] , bx[60][60] , by[60][60] , tx , ty , head[N] , to[M] , val[M] , next[M] , cnt = 1 , s , t , dis[N];char str[60];void add(int x , int y , int z){ to[++cnt] = y , val[cnt] = z , next[cnt] = head[x] , head[x] = cnt; to[++cnt] = x , val[cnt] = 0 , next[cnt] = head[y] , head[y] = cnt;}bool bfs(){ int x , i; memset(dis , 0 , sizeof(dis)); while(!q.empty()) q.pop(); dis[s] = 1 , q.push(s); while(!q.empty()) { x = q.front() , q.pop(); for(i = head[x] ; i ; i = next[i]) { if(val[i] && !dis[to[i]]) { dis[to[i]] = dis[x] + 1; if(to[i] == t) return 1; q.push(to[i]); } } } return 0;}int dinic(int x , int low){ if(x == t) return low; int temp = low , i , k; for(i = head[x] ; i ; i = next[i]) { if(val[i] && dis[to[i]] == dis[x] + 1) { k = dinic(to[i] , min(temp , val[i])); if(!k) dis[to[i]] = 0; val[i] -= k , val[i ^ 1] += k; if(!(temp -= k)) break; } } return low - temp;}int main(){ int n , m , i , j , ans = 0; scanf("%d%d" , &n , &m); for(i = 1 ; i <= n ; i ++ ) { scanf("%s" , str + 1); for(j = 1 ; j <= m ; j ++ ) map[i][j] = (str[j] == '*' ? 0 : str[j] == 'x' ? 1 : 2); } for(i = 1 ; i <= n ; i ++ ) { tx ++ ; for(j = 1 ; j <= m ; j ++ ) { bx[i][j] = tx; if(map[i][j] == 2) tx ++ ; } } for(j = 1 ; j <= m ; j ++ ) { ty ++ ; for(i = 1 ; i <= n ; i ++ ) { by[i][j] = ty; if(map[i][j] == 2) ty ++ ; } } s = 0 , t = tx + ty + 1; for(i = 1 ; i <= tx ; i ++ ) add(s , i , 1); for(i = 1 ; i <= ty ; i ++ ) add(i + tx , t , 1); for(i = 1 ; i <= n ; i ++ ) for(j = 1 ; j <= m ; j ++ ) if(!map[i][j]) add(bx[i][j] , by[i][j] + tx , 1); while(bfs()) ans += dinic(s , 1 << 30); printf("%d\n" , ans); return 0;}
转载地址：http://jrqji.baihongyu.com/