HDU 3395 Special Fish【最大权匹配】
题意：最大权匹配
分析：最大全匹配
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 const int maxn = 105; 7 const int INF = 2000000000; 8 9 bool Sx[maxn], Sy[maxn]; 10 int Lx[maxn], Ly[maxn]; 11 int Link[maxn]; 12 int W[maxn][maxn]; 13 int n, m; 14 15 bool Find(int i) { 16 Sx[i] = true; 17 for(int j = 1; j <= m; j++) { 18 if(!Sy[j] && Lx[i] + Ly[j] == W[i][j]) { 19 Sy[j] = true; 20 if(Link[j] == 0 || Find(Link[j])) { 21 Link[j] = i; 22 return true; 23 } 24 } 25 } 26 return false; 27 } 28 29 int KM() { 30 for(int i = 1; i <= n;i++) { 31 Lx[i] = 0; 32 for(int j = 1; j <= m; j++) { 33 Lx[i] = max(Lx[i], W[i][j]); 34 } 35 } 36 memset(Ly, 0, sizeof(Ly)); 37 memset(Link, 0, sizeof(Link)); 38 for(int v = 1; v <= n; v++) { 39 memset(Sx, 0, sizeof(Sx)); 40 memset(Sy, 0, sizeof(Sy)); 41 while(1) { 42 if(Find(v)) break; 43 int dmin = INF; 44 for(int i = 1; i <= n; i++) { 45 if(Sx[i]) { 46 for(int j = 1; j <= m; j++) { 47 if(!Sy[j] && Lx[i] + Ly[j] - W[i][j] < dmin) { 48 dmin = Lx[i] + Ly[j] - W[i][j]; 49 } 50 } 51 } 52 } 53 for(int i = 1; i <= n; i++) { 54 if(Sx[i]) { 55 Lx[i] -= dmin; 56 Sx[i] = 0; 57 } 58 } 59 for(int i = 1; i <= m; i++) { 60 if(Sy[i]) { 61 Ly[i] += dmin; 62 Sy[i] = 0; 63 } 64 } 65 } 66 } 67 int sum = 0; 68 for(int i = 1; i <= n; i++) { 69 sum += W[Link[i]][i]; 70 } 71 return sum; 72 } 73 74 int val[maxn]; 75 char mat[maxn][maxn]; 76 int main() { 77 while(scanf("%d",&n) && n) { 78 memset(W, 0, sizeof(W)); 79 for(int i = 1; i <= n; i++) { 80 scanf("%d",&val[i]); 81 } 82 for(int i = 1; i <= n; i++) { 83 scanf("%s",mat[i]); 84 } 85 for(int i = 1; i <= n; i++) { 86 for(int j = 0; j < n; j++) { 87 if(i == j + 1) continue; 88 if(mat[i][j] == '1') { 89 W[i][j + 1] = val[i] ^ val[j + 1]; 90 } 91 } 92 } 93 m = n; 94 printf("%d\n",KM()); 95 } 96 return 0; 97 }
