hdu KM匹配题集
模板 2255
1 /*Author :usedrose */ 2 /*Created Time :2015/8/30 1:20:31*/ 3 /*File Name :2.cpp*/ 4 #pragma comment(linker, "/STACK:1024000000,1024000000") 5 #include <cstdio> 6 #include <iostream> 7 #include <algorithm> 8 #include <sstream> 9 #include <cstdlib> 10 #include <cstring> 11 #include <climits> 12 #include <vector> 13 #include <string> 14 #include <ctime> 15 #include <cmath> 16 #include <deque> 17 #include <queue> 18 #include <stack> 19 #include <set> 20 #include <map> 21 #define eps 1e-8 22 #define pi acos(-1.0) 23 #define OK cout << "ok" << endl; 24 #define o(a) cout << #a << " = " << a << endl 25 #define o1(a,b) cout << #a << " = " << a << " " << #b << " = " << b << endl 26 using namespace std; 27 typedef long long LL; 28 29 const int N = 310; 30 const int INF = 0x3f3f3f3f; 31 int nx,ny;//两边的点数 32 int g[N][N];//二分图描述 33 int linker[N],lx[N],ly[N];//y中各点匹配状态, x,y中的点标号 34 int slack[N]; 35 bool visx[N],visy[N]; 36 bool DFS(int x) 37 { 38 visx[x] = true; 39 for(int y = 0; y < ny; y++) { 40 if(visy[y])continue; 41 int tmp = lx[x] + ly[y] - g[x][y]; 42 if(tmp == 0) { 43 visy[y] = true; 44 if(linker[y] == -1 || DFS(linker[y])) { 45 linker[y] = x; 46 return true; 47 } 48 } 49 else if(slack[y] > tmp) 50 slack[y] = tmp; 51 } 52 return false; 53 } 54 55 int KM() 56 { 57 memset(linker,-1,sizeof(linker)); 58 memset(ly,0,sizeof(ly)); 59 for(int i = 0;i < nx;i++) { 60 lx[i] = -INF; 61 for(int j = 0;j < ny;j++) 62 if(g[i][j] > lx[i]) 63 lx[i] = g[i][j]; 64 } 65 for(int x = 0;x < nx;x++) { 66 for(int i = 0;i < ny;i++) 67 slack[i] = INF; 68 while(true) { 69 memset(visx,false,sizeof(visx)); 70 memset(visy,false,sizeof(visy)); 71 if(DFS(x))break; 72 int d = INF; 73 for(int i = 0;i < ny;i++) 74 if(!visy[i] && d > slack[i]) 75 d = slack[i]; 76 for(int i = 0;i < nx;i++) 77 if(visx[i]) 78 lx[i] -= d; 79 for(int i = 0;i < ny;i++) { 80 if(visy[i])ly[i] += d; 81 else slack[i] -= d; 82 } 83 } 84 } 85 int res = 0; 86 for(int i = 0;i < ny;i++) 87 if(linker[i] != -1) 88 res += g[linker[i]][i]; 89 return res; 90 } 91 92 int main() 93 { 94 int n; 95 while(scanf("%d",&n) == 1) { 96 for(int i = 0;i < n;i++) 97 for(int j = 0;j < n;j++) 98 scanf("%d",&g[i][j]); 99 nx = ny = n; 100 printf("%d\n",KM()); 101 } 102 return 0; 103 }
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