C++ CE G++ AC什么鬼...
这题虽说是网络流 但是可以用之前的KM最优匹配做
会的话还是比较好写的
这里也发现了最大流/费用流更适合离散图 匈牙利/KM更适合稀疏图
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 typedef double D; 13 #define eps 1e-8 14 ll read() { 15 ll as = 0,fu = 1; 16 char c = getchar(); 17 while(c < '0' || c > '9') { 18 if(c == '-') fu = -1; 19 c = getchar(); 20 } 21 while(c >= '0' && c <= '9') { 22 as = as * 10 + c - '0'; 23 c = getchar(); 24 } 25 return as * fu; 26 } 27 //head 28 const int N = 1005; 29 int n,m; 30 char cmd[N]; 31 32 struct node { 33 int x,y; 34 }a[N],b[N]; 35 int top1,top2; 36 int dis(node a,node b) { 37 return abs(a.x-b.x) + abs(a.y-b.y); 38 } 39 void input() { 40 top1 = top2 = 0; 41 rep(i,1,n) { 42 scanf("%s",cmd+1); 43 rep(j,1,m) { 44 if(cmd[j] == 'm') a[++top1] = (node){i,j}; 45 if(cmd[j] == 'H') b[++top2] = (node){i,j}; 46 } 47 } 48 } 49 50 int v[N][N],ans; 51 int lx[N],ly[N]; 52 int match[N]; 53 bool s[N],t[N]; 54 bool dfs(int x) { 55 s[x] = 1; 56 rep(y,1,n) { 57 if(lx[x] + ly[y] == v[x][y] && !t[y]) { 58 t[y] = 1; 59 if(!match[y] || dfs(match[y])) { 60 match[y] = x; 61 return 1; 62 } 63 } 64 } 65 return 0; 66 } 67 68 void update() { 69 int flw = inf; 70 rep(x,1,n) if(s[x]) { 71 rep(y,1,n) if(!t[y]) { 72 flw = min(flw,lx[x] + ly[y] - v[x][y]); 73 } 74 } 75 rep(i,1,n) { 76 if(s[i]) lx[i] -= flw; 77 if(t[i]) ly[i] += flw; 78 } 79 } 80 81 void KM() { 82 rep(i,1,n) { 83 match[i] = lx[i] = ly[i] = 0; 84 rep(j,1,n) lx[i] = max(lx[i],v[i][j]); 85 } 86 rep(i,1,n) { 87 while(1) { 88 rep(j,1,n) s[j] = t[j] = 0; 89 if(dfs(i)) break; 90 update(); 91 } 92 } 93 } 94 95 int main() { 96 while(~scanf("%d%d",&n,&m) && n && m) { 97 input(),n = top1; 98 rep(i,1,n) rep(j,1,n) v[i][j] = -dis(a[i],b[j]); 99 KM(),ans = 0; 100 rep(i,1,n) ans += lx[i] + ly[i]; 101 printf("%d ",-ans); 102 } 103 return 0; 104 }