题目大意：给定一个N*N矩阵，每个格子(i,j)有正权值w(i,j)，要求你为每一行r、每一列c确定一个数l(r)、l(c)，要求对任一格子(i,j)有l(r) + l(c) >= w(i, j),要求所有l之和尽量小
题解：KM裸题，不学KM可能会死很惨
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <map> 9 #include <string> 10 #include <cmath> 11 #define min(a, b) ((a) < (b) ? (a) : (b)) 12 #define max(a, b) ((a) > (b) ? (a) : (b)) 13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 14 template<class T> 15 inline void swap(T &a, T &b) 16 { 17 T tmp = a;a = b;b = tmp; 18 } 19 inline void read(int &x) 20 { 21 x = 0;char ch = getchar(), c = ch; 22 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 23 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 24 if(c == '-') x = -x; 25 } 26 const int INF = 0x3f3f3f3f; 27 const int MAXN = 500 + 10; 28 int g[MAXN][MAXN], n1, n2, lab1[MAXN], lab2[MAXN], lk1[MAXN], lk2[MAXN], pre[MAXN], vis[MAXN], sla[MAXN]; 29 void cal(int x) 30 { 31 memset(vis, 0, sizeof(vis)), memset(pre, 0, sizeof(pre)), memset(sla, 0x3f, sizeof(sla)), vis[0] = 1; 32 int y; 33 do 34 { 35 y = 0; 36 for(int i = 1;i <= n2;++ i) 37 { 38 if(vis[i]) continue; 39 if(lab1[x] + lab2[i] - g[x][i] < sla[i]) sla[i] = lab1[x] + lab2[i] - g[x][i], pre[i] = x; 40 if(sla[i] < sla[y]) y = i; 41 } 42 int d = sla[y]; 43 for(int i = 1;i <= n1;++ i) if(vis[lk1[i]]) lab1[i] -= d; 44 for(int i = 1;i <= n2;++ i) if(vis[i]) lab2[i] += d; else sla[i] -= d; 45 vis[y] = 1; 46 }while(x = lk2[y]); 47 for(;y;swap(y, lk1[lk2[y] = pre[y]])); 48 } 49 int KM() 50 { 51 for(int i = 1;i <= n1;++ i) cal(i); int ans = 0; 52 for(int i = 1;i <= n1;++ i) ans += g[i][lk1[i]]; return ans; 53 } 54 int main() 55 { 56 while(scanf("%d", &n1) != EOF) 57 { 58 n2 = n1;memset(g, 0, sizeof(g)), memset(lab1, 0, sizeof(lab1)), memset(lab2, 0, sizeof(lab2)), memset(lk1, 0, sizeof(lk1)), memset(lk2, 0, sizeof(lk2)); 59 for(int i = 1;i <= n1;++ i) 60 for(int j = 1;j <= n2;++ j) 61 read(g[i][j]), lab1[i] = max(lab1[i], g[i][j]); 62 int ans = KM(); 63 for(int i = 1;i < n1;++ i) printf("%d ", lab1[i]); printf("%d ", lab1[n1]); 64 for(int i = 1;i < n2;++ i) printf("%d ", lab2[i]); printf("%d ", lab2[n2]); 65 printf("%d ", ans); 66 } 67 return 0; 68 }