脑残题
建图都懒得说了
——代码
1 #include <queue> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #define N 1000001 6 #define min(x, y) ((x) < (y) ? (x) : (y)) 7 8 int n, cnt, s, t; 9 int a[101][101], dis[N], pre[N]; 10 int head[N], to[N << 1], val[N << 1], cost[N << 1], next[N << 1]; 11 bool vis[N]; 12 13 inline int read() 14 { 15 int x = 0, f = 1; 16 char ch = getchar(); 17 for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; 18 for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; 19 return x * f; 20 } 21 22 inline void add(int x, int y, int z, int c) 23 { 24 to[cnt] = y; 25 val[cnt] = z; 26 cost[cnt] = c; 27 next[cnt] = head[x]; 28 head[x] = cnt++; 29 } 30 31 inline bool spfa() 32 { 33 int i, u, v; 34 std::queue <int> q; 35 memset(vis, 0, sizeof(vis)); 36 memset(pre, -1, sizeof(pre)); 37 memset(dis, 127 / 3, sizeof(dis)); 38 q.push(s); 39 dis[s] = 0; 40 while(!q.empty()) 41 { 42 u = q.front(), q.pop(); 43 vis[u] = 0; 44 for(i = head[u]; i ^ -1; i = next[i]) 45 { 46 v = to[i]; 47 if(val[i] && dis[v] > dis[u] + cost[i]) 48 { 49 dis[v] = dis[u] + cost[i]; 50 pre[v] = i; 51 if(!vis[v]) 52 { 53 q.push(v); 54 vis[v] = 1; 55 } 56 } 57 } 58 } 59 return pre[t] ^ -1; 60 } 61 62 inline int dinic() 63 { 64 int i, d, sum = 0; 65 while(spfa()) 66 { 67 d = 1e9; 68 for(i = pre[t]; i ^ -1; i = pre[to[i ^ 1]]) d = min(d, val[i]); 69 for(i = pre[t]; i ^ -1; i = pre[to[i ^ 1]]) 70 { 71 val[i] -= d; 72 val[i ^ 1] += d; 73 } 74 sum += dis[t] * d; 75 } 76 return sum; 77 } 78 79 int main() 80 { 81 int i, j, x; 82 n = read(); 83 s = 0, t = n + n + 1; 84 memset(head, -1, sizeof(head)); 85 for(i = 1; i <= n; i++) 86 { 87 add(s, i, 1, 0); 88 add(i, s, 0, 0); 89 add(i + n, t, 1, 0); 90 add(t, i + n, 0, 0); 91 for(j = 1; j <= n; j++) 92 { 93 a[i][j] = read(); 94 add(i, j + n, 1, a[i][j]); 95 add(j + n, i, 0, -a[i][j]); 96 } 97 } 98 printf("%d ", dinic()); 99 cnt = 0; 100 memset(head, -1, sizeof(head)); 101 for(i = 1; i <= n; i++) 102 { 103 add(s, i, 1, 0); 104 add(i, s, 0, 0); 105 add(i + n, t, 1, 0); 106 add(t, i + n, 0, 0); 107 for(j = 1; j <= n; j++) 108 { 109 add(i, j + n, 1, -a[i][j]); 110 add(j + n, i, 0, a[i][j]); 111 } 112 } 113 printf("%d ", -dinic()); 114 return 0; 115 }