qyy开始练习网络流啦 ， 啊 ，蒟蒻只会套版 ，很裸的题 ， 我连题都不想发了 ，可以参考我的代码（虽然我也是看的别人的
1 #include <iostream> 2 #include <cstring> 3 #include <vector> 4 #include <cstdio> 5 #include <queue> 6 const int N = 1000 + 11 , M = 300000 + 11 ; 7 using namespace std ; 8 int n , m ; 9 int tot , head[N] ; 10 long long ans; 11 struct id 12 { 13 int to , nxt ; 14 long long val ; 15 } edge[M<<1] ; 16 queue< int > Q; 17 int dis[N] ,cur[N]; 18 19 inline void Init( ) 20 { 21 freopen( "NSOOJ10477.in" , "r" , stdin ) ; 22 freopen( "NSOOJ10477.out" , "w" , stdout ) ; 23 } 24 25 26 int read( ) 27 { 28 char ch = getchar( ) ; int k = 1 , ret = 0 ; 29 while( ch > '9' || ch < '0' ) { if( ch == '-' ) k = -1 ; ch = getchar( ) ; } 30 while( ch <= '9' && ch >= '0' ) ret = ret * 10 + ch - '0' , ch = getchar( ) ; 31 return ret * k ; 32 } 33 34 35 void add( int u , int v , int c ) 36 { 37 edge[++tot].to = v , edge[tot].nxt = head[u] ; 38 edge[tot].val = c , head[u] = tot ; 39 } 40 41 42 void input( ) 43 { 44 n = read( ) , m = read( ) ; 45 memset( head , -1 , sizeof( head ) ) ; 46 tot = 1 ; 47 for( int x = 1 ; x <= m ; ++x ) 48 { 49 int a, b , c ; 50 a = read( ) , b = read( ) , c = read( ) ; 51 add( a, b , c ) ; 52 add( b , a , 0 ) ; 53 } 54 } 55 56 bool bfs( ) 57 { 58 memset( dis , -1 , sizeof(dis) ) ; 59 Q.push( 1 ) ; dis[1] = 0 ; 60 while( !Q.empty( ) ) 61 { 62 int u = Q.front( ) ; Q.pop( ) ; 63 for( int i = head[u] ; ~i ; i = edge[i].nxt ) 64 { 65 66 int v = edge[i].to ; 67 if( dis[v] < 0 && edge[i].val > 0 ) 68 { 69 dis[v] = dis[u] + 1 ; Q.push( v ) ; 70 } 71 72 } 73 } 74 // cout<<endl; 75 if( dis[n] > 0 ) return true ; 76 return false ; 77 } 78 79 80 long long int dfs( int u , int inf ) 81 { 82 long long int ans = 0ll , cost = 0ll; 83 if( u == n ) return inf ; 84 for( int i = cur[u] ; ~i ; i = edge[i].nxt ) 85 { 86 int v = edge[i].to ; 87 if( dis[v] != dis[u] + 1 ) continue ; 88 ans = dfs( v , min( inf - cost , edge[i].val ) ) ; 89 edge[i].val -= ans ; 90 edge[i^1].val += ans ; 91 if( edge[i].val ) cur[u] = i ; 92 cost += ans ; 93 if( cost == inf ) return inf ; 94 } 95 if( !cost ) dis[u] = -1 ; 96 return cost ; 97 } 98 99 100 void sov( ) 101 { 102 ans = 0 ; 103 while( bfs( ) ) 104 { 105 int now ; 106 for( int i = 1 ; i <= n ; ++i ) cur[i] = head[i] ; 107 ans += dfs( 1 , 1<<30 ) ; 108 } 109 printf( "%lld" , ans ) ; 110 } 111 112 113 int main( ) 114 { 115 // Init( ) ; 116 input( ) ; 117 sov( ) ; 118 // fclose( stdin ) ; 119 // fclose( stdout ) ; 120 return 0 ; 121 }
重新贴板，和上面的程序几乎没什么区别，码风固定了，值得开心。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <queue> 5 const int N = 1003, M = 3e5 + 7, inf = 1 << 30; 6 using namespace std; 7 int n, m, head[N], cnt, dis[N],cur[N]; 8 queue<int>Q; 9 struct edge 10 { 11 int nxt,to,vi; 12 } E[M<<1]; 13 14 void add(int u,int v,int vi) 15 { 16 E[++cnt] = (edge){head[u],v,vi}; head[u] = cnt; 17 E[++cnt] = (edge){head[v],u,0}; head[v] = cnt; 18 } 19 20 void Init() 21 { 22 scanf("%d%d",&n,&m); cnt = -1; 23 int u,v,vi; 24 memset(head,-1,sizeof(head)); 25 for(int i = 1; i <= m; ++i) 26 { 27 scanf("%d%d%d",&u,&v,&vi); 28 add(u,v,vi); 29 } 30 } 31 32 bool bfs() 33 { 34 memset(dis,-1,sizeof(dis)); 35 Q.push(1); dis[1] = 0; 36 while(!Q.empty()) 37 { 38 int u = Q.front(); Q.pop(); 39 for(int i = head[u]; ~i; i = E[i].nxt) 40 { 41 int v = E[i].to; 42 if(dis[v] == -1 && E[i].vi > 0) dis[v] = dis[u] + 1, Q.push(v); 43 } 44 } 45 return (dis[n] > 0); 46 } 47 48 int dfs(int u,int t,int ff) 49 { 50 int ret = 0, cost = 0; 51 if(u == t) return ff; 52 for(int i = cur[u] ; ~i; i = E[i].nxt) 53 { 54 int v = E[i].to; 55 if(dis[v] != dis[u] + 1) continue; 56 cost = dfs(v,t,min(ff-ret,E[i].vi)); 57 E[i].vi -= cost, E[i^1].vi += cost; ret += cost; 58 if(E[i].vi) cur[u] = i; 59 if(ret == ff) return ff; 60 } 61 if(!ret) dis[u] = -1; 62 return ret; 63 } 64 65 int max_flow(int s,int t) 66 { 67 int ret = 0; 68 while(bfs()) 69 { 70 for(int i = 1; i <= n; ++i) cur[i] = head[i]; 71 ret += dfs(s,t,inf); 72 } 73 return ret; 74 } 75 76 void Solve() 77 { 78 int ans = max_flow(1,n); 79 printf("%d ",ans); 80 } 81 82 int main() 83 { 84 Init(); 85 Solve(); 86 87 return 0; 88 }