没什么好说的，用spfa来找增广路。
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 5 using namespace std; 6 7 const int INF = 1 << 26; 8 int n, m, s, t, cnt; 9 int to[100001], val[100001], next[100001], cost[100001], head[5001], pre[5001], path[5001], dis[5001]; 10 //pre记录前一个节点，path记录边 11 bool vis[5001]; 12 13 inline int read()//读入优化 14 { 15 int x = 0, f = 1; 16 char ch = getchar(); 17 while(ch < '0' || ch > '9') 18 { 19 if(ch == '-') f = -1; 20 ch = getchar(); 21 } 22 while(ch >= '0' && ch <= '9') 23 { 24 x = x * 10 + ch - '0'; 25 ch = getchar(); 26 } 27 return x * f; 28 } 29 30 inline void add(int a, int b, int c, int d) 31 { 32 to[cnt] = b; 33 val[cnt] = c; 34 cost[cnt] = d; 35 next[cnt] = head[a]; 36 head[a] = cnt++; 37 } 38 39 inline bool spfa() 40 { 41 int u, i, v; 42 memset(vis, 0, sizeof(vis)); 43 memset(pre, -1, sizeof(pre)); 44 for(i = 1; i <= n; i++) dis[i] = INF; 45 dis[s] = 0; 46 queue <int> q; 47 q.push(s); 48 vis[s] = 1; 49 while(!q.empty()) 50 { 51 u = q.front(); 52 vis[u] = 0; 53 q.pop(); 54 for(i = head[u]; i != -1; i = next[i]) 55 { 56 v = to[i]; 57 if(val[i] > 0 && dis[v] > dis[u] + cost[i]) 58 { 59 dis[v] = dis[u] + cost[i]; 60 pre[v] = u; 61 path[v] = i; 62 if(!vis[v]) 63 { 64 q.push(v); 65 vis[v] = 1; 66 } 67 } 68 } 69 } 70 return pre[t] != -1; 71 } 72 73 int main() 74 { 75 int i, j, a, b, c, d, money = 0, flow = 0, f; 76 n = read(); 77 m = read(); 78 s = read(); 79 t = read(); 80 memset(head, -1, sizeof(head)); 81 for(i = 1; i <= m; i++) 82 { 83 a = read(); 84 b = read(); 85 c = read(); 86 d = read(); 87 add(a, b, c, d); 88 add(b, a, 0, -d); 89 } 90 while(spfa()) 91 { 92 f = INF; 93 for(i = t; i != s; i = pre[i]) f = min(f, val[path[i]]); 94 flow += f; 95 money += dis[t] * f; 96 for(i = t; i != s; i = pre[i]) 97 { 98 val[path[i]] -= f; 99 val[path[i] ^ 1] += f; 100 } 101 } 102 printf("%d %d", flow, money); 103 return 0; 104 }
但是spfa对于稠密图的处理效果不是很好，所以改用dijkstra+stl堆优化，至少快了些。。
——代码
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #define Heap pair<int, int> 5 6 using namespace std; 7 8 const int INF = 1 << 26; 9 int n, m, s, t, cnt; 10 int to[100001], val[100001], next[100001], cost[100001], head[5001], pre[5001], path[5001], dis[5001]; 11 //pre记录前一个节点，path记录边 12 13 inline int read()//读入优化 14 { 15 int x = 0, f = 1; 16 char ch = getchar(); 17 while(ch < '0' || ch > '9') 18 { 19 if(ch == '-') f = -1; 20 ch = getchar(); 21 } 22 while(ch >= '0' && ch <= '9') 23 { 24 x = x * 10 + ch - '0'; 25 ch = getchar(); 26 } 27 return x * f; 28 } 29 30 inline void add(int a, int b, int c, int d) 31 { 32 to[cnt] = b; 33 val[cnt] = c; 34 cost[cnt] = d; 35 next[cnt] = head[a]; 36 head[a] = cnt++; 37 } 38 39 bool Dijkstra() 40 { 41 int i, v; 42 Heap u; 43 memset(pre, -1, sizeof(pre)); 44 for(i = 1; i <= n; i++) dis[i] = INF; 45 dis[s] = 0; 46 priority_queue <Heap, vector <Heap>, greater <Heap> > q; 47 q.push(make_pair(0, s));//前一个表示当前节点到起点的距离，后一个是当前节点 48 while(!q.empty()) 49 { 50 u = q.top(); 51 q.pop(); 52 if(u.first != dis[u.second]) continue; 53 for(i = head[u.second]; i != -1; i = next[i]) 54 { 55 v = to[i]; 56 if(val[i] > 0 && dis[v] > dis[u.second] + cost[i]) 57 { 58 dis[v] = dis[u.second] + cost[i]; 59 pre[v] = u.second; 60 path[v] = i; 61 q.push(make_pair(dis[v], v)); 62 } 63 } 64 } 65 return pre[t] != -1; 66 } 67 68 int main() 69 { 70 int i, j, a, b, c, d, money = 0, flow = 0, f; 71 n = read(); 72 m = read(); 73 s = read(); 74 t = read(); 75 memset(head, -1, sizeof(head)); 76 for(i = 1; i <= m; i++) 77 { 78 a = read(); 79 b = read(); 80 c = read(); 81 d = read(); 82 add(a, b, c, d); 83 add(b, a, 0, -d); 84 } 85 while(Dijkstra()) 86 { 87 f = INF; 88 for(i = t; i != s; i = pre[i]) f = min(f, val[path[i]]); 89 flow += f; 90 money += dis[t] * f; 91 for(i = t; i != s; i = pre[i]) 92 { 93 val[path[i]] -= f; 94 val[path[i] ^ 1] += f; 95 } 96 } 97 printf("%d %d", flow, money); 98 return 0; 99 }
洛谷改了数据之后dijk居然T了。。