[BZOJ1266]-[AHOI2006]上学路线route-SPFA+最小割

说在前面

感觉早期的题都是在考算法= =
不过至少比那种把小学奥数拿到NOIP凑数的好hhhhh

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题目

BZOJ1266传送门

题面

由于题面实在是太长了，Izumi并不想把一大段文字直接粘过来
这题不是权限题，可以进去看题面

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解法

读完题大概都知道要干什么了吧=w=
要删掉一些边使得最短路变长，那就必须割掉所有最短路的某条边。
在此基础上还要费用最小，最小割->最大流->模板题

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下面是自带大常数的代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int N , M , head[505] , Whead[505] , tp , tw = 1 ;struct Path{    int pre , to , len , flow ;}p[250000] , w[250000] ;inline void In( int t1 , int t2 , int t3 , int t4 ){    p[++tp].pre = head[t1] ;    p[ head[t1] = tp ].to = t2 ;    p[tp].len = t3 ;    p[tp].flow = t4 ;}inline void WIn( int t1 , int t2 , int t3 ){    w[++tw].pre = Whead[t1] ;    w[ Whead[t1] = tw ].to = t2 ;    w[tw].flow = t3 ;}bool inque[505] ;int dis[505] , que[250005] , fr , ba ;void SPFA(){    fr = 125001 , ba = 125000 ;    memset( dis , 0x3f , sizeof( dis ) ) ;    dis[1] = 0 ; que[++ba] = 1 ; inque[1] = true ;    while( ba >= fr ){        int u = que[fr++] ; inque[u] = false ;        for( int i = head[u] ; i ; i = p[i].pre ){            int v = p[i].to ;            if( dis[v] > dis[u] + p[i].len ){                dis[v] = dis[u] + p[i].len ;                if( !inque[v] ){                    if( dis[v] < dis[ que[fr] ] ) que[--fr] = v ;                    else que[++ba] = v ;                    inque[v] = true ;                }            }        }    }    printf( "%d\n" , dis[N] ) ;}void build(){    for( int u = 1 ; u <= N ; u ++ )        for( int i = head[u] ; i ; i = p[i].pre )            if( dis[u] + p[i].len == dis[ p[i].to ] ){                WIn( u , p[i].to , p[i].flow ) ;                WIn( p[i].to , u , 0 ) ;            }}bool Bfs(){    memset( dis , -1 , sizeof( dis ) ) ;    dis[1] = 0 ;    fr = 1 , ba = 0 ; que[++ba] = 1 ;    while( fr <= ba ){        int u = que[fr++] ;        for( int i = Whead[u] ; i ; i = w[i].pre ){            int v = w[i].to ;            if( dis[v] != -1 || !w[i].flow ) continue ;            dis[v] = dis[u] + 1 ;            que[++ba] = v ;        }    }    return dis[N] != -1 ;}int dfs( int u , int flow ){    if( u == N || !flow ) return flow ;    int totf = 0 ;    for( int i = Whead[u] ; i ; i = w[i].pre ){        int v = w[i].to ;        if( dis[v] != dis[u] + 1 || !w[i].flow ) continue ;        int nowf = dfs( v , min( flow , w[i].flow ) ) ;        if( nowf ){            totf += nowf ;            flow -= nowf ;            w[i].flow -= nowf ;            w[i^1].flow += nowf ;        }    }    if( !totf ) dis[u] = -1 ;    return totf ;}void solve(){    SPFA() ;    build() ;    int ans = 0 ;    while( Bfs() )        ans += dfs( 1 , 0x3f3f3f3f ) ;    printf( "%d" , ans ) ;}inline int read_(){    int rt = 0 ;    char ch = getchar() ;    while( ch < '0' || ch > '9' ) ch = getchar() ;    while( ch >='0' && ch <='9' ) rt = (rt<<1) + (rt<<3) + ch - '0' , ch = getchar() ;    return rt ;}int main(){    scanf( "%d%d" , &N , &M ) ;    for( register int i = 1 , u , v , t , c ; i <= M ; i ++ ){        u = read_() , v = read_() , t = read_() , c = read_() ;        In( u , v , t , c ) ;        In( v , u , t , c ) ;    }    solve() ;}