一道水題WA了這么多次真是....

統考終于完 ( 掛 ) 了...可以好好寫題了...

先floyd跑出各個點的最短路 , 然后二分答案 m , 再建圖.

每個 farm 拆成一個 cow 點和一個 shelter 點, 然后對于每個 farm x : S -> cow( x ) = cow( x ) 數量 , shelter( x ) -> T = shelter( x ) 容量 ; 對于每個dist( u , v ) <= m 的 cow( u ) -> shelter( v ) ?= +oo , 然后跑最大流 , 假如滿流就可行

最近我真是積極寫題解...雖說都是水題..攢RP...

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#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#define rep( i , n ) for( int i = 0 ; i < n ; ++i )#define clr( x , c ) memset( x , c , sizeof( x ) )#define Rep( i , n ) for( int i = 1 ; i <= n ; ++i )using namespace std;typedef long long ll;const int maxn = 410;const int maxm = 41000;const int inf = 1e8;const ll oo = ll( 1e15 );inline int read() {int ans = 0;char c = getchar();for( ; ! isdigit( c ) ; c = getchar() );for( ; isdigit( c ) ; c = getchar() ) ? ?ans = ans * 10 + c - '0';return ans;}struct edge {int to , cap;edge *next , *rev;};edge* pt , E[ maxm << 1 ];edge* head[ maxn ];inline void add( int u , int v , int d ) {pt -> to = v;pt -> cap = d;pt -> next = head[ u ];head[ u ] = pt++;}inline void add_edge( int u , int v , int d ) {add( u , v , d );add( v , u , 0 );head[ u ] -> rev = head[ v ];head[ v ] -> rev = head[ u ];}inline void edge_init() {clr( head , 0 );pt = E;}edge *p[ maxn ] , *cur[ maxn ];int cnt[ maxn ] , h[ maxn ];int N , S , T , COW = 0;bool check() { // max flowrep( i , N ) cur[ i ] = head[ i ];clr( h , 0 );clr( cnt , 0 );cnt[ 0 ] = N;edge* e;int flow = 0 , x = S , A = inf;while( h[ S ] < N ) {for( e = cur[ x ] ; e ; e = e -> next )if( e -> cap && h[ e -> to ] == h[ x ] - 1 ) break;if( e ) {p[ e -> to ] = cur[ x ] = e;A = min( A , e -> cap );x = e -> to;if( x == T ) {for( ; x != S ; x = p[ x ] -> rev -> to ) {p[ x ] -> cap -= A;p[ x ] -> rev -> cap += A;}flow += A;A = inf;}} else {if( ! --cnt[ h[ x ] ] ) break;h[ x ] = N;for( e = head[ x ] ; e ; e = e -> next ) ? ?if( e -> cap && h[ e -> to ] + 1 < h[ x ] ) {h[ x ] = h[ e -> to ] + 1;cur[ x ] = e; ? ?}cnt[ h[ x ] ]++;if( x != S ) x = p[ x ] -> rev -> to;}}return flow == COW;}ll mp[ maxn ][ maxn ];int a[ maxn ] , b[ maxn ];int n;inline void AE( ll lim ) {rep( i , n ) {add_edge( S , i + 1 , a[ i ] );add_edge( i + 1 + n , T , b[ i ] );}rep( i , n )? ? ?rep( j , n ) if( mp[ i ][ j ] <= lim )? ? ? ? ?add_edge( i + 1 , j + 1 + n , inf );}void init( int m ) {rep( i , n ) ? ?for( int j = i + 1 ; j < n ; j++ ) ? ? ? ?mp[ i ][ j ] = mp[ j ][ i ] = oo;rep( i , n ) mp[ i ][ i ] = 0;while( m-- ) {int u = read() , v = read();ll d = read();u-- , v--;mp[ u ][ v ] = mp[ v ][ u ] = min( d , mp[ u ][ v ] );}rep( k , n ) ? ?rep( i , n ) if( mp[ i ][ k ] != oo ) ? ? ? ?rep( j , n ) if( mp[ k ][ j ] != oo ) ? ? ? ? ? ?mp[ i ][ j ] = min( mp[ i ][ k ] + mp[ k ][ j ] , mp[ i ][ j ] );}ll BS() {ll L = 0 , R = oo - 1 , ans = -1;while( L <= R ) {edge_init();ll m = ( L + R ) >> 1;AE( m );if( check() )R = ( ans = m ) - 1;else ? ?L = m + 1;}return ans;}int main() {freopen( "test.in" , "r" , stdin );int p;cin >> n >> p;S = 0 , T = ( n << 1 ) ^ 1 , N = T + 1;rep( i , n ) {COW += ( a[ i ] = read() );b[ i ] = read();}init( p );cout << BS() << "\n";return 0;}

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1738: [Usaco2005 mar]Ombrophobic Bovines 發抖的牛

Time Limit:?5 Sec??Memory Limit:?64 MB
Submit:?206??Solved:?91
[Submit][Status][Discuss]

Description

FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter. The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction. Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse. Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

????約翰的牛們非常害怕淋雨，那會使他們瑟瑟發抖．他們打算安裝一個下雨報警器，并且安排了一個撤退計劃．他們需要計算最少的讓所有牛進入雨棚的時間．????牛們在農場的F(1≤F≤200)個田地上吃草．有P(1≤P≤1500)條雙向路連接著這些田地．路很寬，無限量的?？梢酝ㄟ^．田地上有雨棚，雨棚有一定的容量，牛們可以瞬間從這塊田地進入這塊田地上的雨棚????請計算最少的時間，讓每只牛都進入雨棚．

Input

* Line 1: Two space-separated integers: F and P

?* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i. * Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

????第1行：兩個整數F和P;????第2到F+1行：第i+l行有兩個整數描述第i個田地，第一個表示田地上的牛數，第二個表示田地上的雨棚容量．兩個整數都在0和1000之間．????第F+2到F+P+I行：每行三個整數描述一條路，分別是起點終點，及通過這條路所需的時間（在1和10^9之間）．

Output

* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".

????一個整數，表示最少的時間．如果無法使牛們全部進入雨棚，輸出-1.

Sample Input

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output

110

1號田的7只牛中，2只牛直接進入1號田的雨棚，4只牛進入1號田的雨棚，1只進入3號田的雨棚，加入其他的由3號田來的牛們．所有的牛都能在110單位時間內到達要去的雨棚．

HINT

Source

Gold

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轉載于:https://www.cnblogs.com/JSZX11556/p/4621241.html

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