[bzoj1834][ZJOI2010] 网络扩容 最大流 费用流

[ZJOI2010]network 网络扩容

Description

给定一张有向图，每条边都有一个容量C和一个扩容费用W。这里扩容费用是指将容量扩大1所需的费用。求： 1、 在不扩容的情况下，1到N的最大流； 2、 将1到N的最大流增加K所需的最小扩容费用。
Input

输入文件的第一行包含三个整数N,M,K，表示有向图的点数、边数以及所需要增加的流量。 接下来的M行每行包含四个整数u,v,C,W，表示一条从u到v，容量为C，扩容费用为W的边。
Output

输出文件一行包含两个整数，分别表示问题1和问题2的答案。
Sample Input

5 8 2

1 2 5 8

2 5 9 9

5 1 6 2

5 1 1 8

1 2 8 7

2 5 4 9

1 2 1 1

1 4 2 1

Sample Output

13 19

30%的数据中，N<=100

100%的数据中，N<=1000，M<=5000，K<=10

我不知道为什么原始对偶挂了，进了死循环
所以学了普通费用流

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int inf = 0x7fffffff;const int N = 1005;const int M = 500000 + 5;int n,m,k,ans=0,T,S;int h[N],last[N],q[N],cnt=1,d[N],from[N];bool inq[N];template <class T>inline bool readIn(T &x)  {    T flag = 1;  char ch;    while(!(isdigit(ch = (char) getchar())) && ch != EOF)  if( ch == '-' )  flag = -1;    if(ch == EOF)  return false;    for(x = ch - 48; isdigit(ch = (char) getchar()); x = (x << 1) + (x << 3) + ch - 48);    x *= flag;    return true;}struct Edge{    int to,v,c,next,from,c1;}e[M];void insert( int u, int v, int w, int c ){    e[++cnt].next = last[u]; last[u] = cnt; e[cnt].c1 = c; e[cnt].to = v; e[cnt].v = w, e[cnt].from = u;    e[++cnt].next = last[v]; last[v] = cnt; e[cnt].c1 = -c; e[cnt].to = u; e[cnt].v = 0, e[cnt].from = v;}void insert2( int u, int v, int w, int c ){    e[++cnt].next = last[u]; last[u] = cnt; e[cnt].c = c; e[cnt].to = v; e[cnt].v = w, e[cnt].from = u;    e[++cnt].next = last[v]; last[v] = cnt; e[cnt].c = -c; e[cnt].to = u; e[cnt].v = 0, e[cnt].from = v;}void rebuild(){    int yu = cnt;    for( int i = 2; i <= yu; i+=2 ) insert2(e[i].from,e[i].to,inf,e[i].c1);    insert(0,1,k,0); ans = 0;}bool bfs(){    int head = 0, tail = 1;    memset(h,-1,sizeof(h));    q[0] = 1; h[1] = 0;    while( head != tail ){        int now = q[head++]; if( head == n ) head = 0;        for( int i = last[now]; i; i = e[i].next ){            if( e[i].v && h[e[i].to] == -1 ){                h[e[i].to] = h[now]+1;                q[tail++] = e[i].to; if( tail == n ) tail = 0;            }        }    }    return h[n] != -1;}int dfs( int x, int f ){    if( x == n ) return f;    int w,used = 0;    for( int i = last[x]; i; i = e[i].next )        if( e[i].v && h[e[i].to] == h[x]+1 ){            w = dfs( e[i].to, min(f-used,e[i].v) );            e[i].v -= w; e[i^1].v += w; used += w;            if( used == f ) return f;        }    if( !used ) h[x] = -1;    return used;}void dinic(){    while(bfs()){        ans += dfs(1,inf);    }}bool spfa(){    for( int i = 1; i <= T; i++ ) d[i] = inf;    int head = 0, tail = 1;    q[0] = S; d[S] = 0; inq[S] = 1;    while( head != tail ){        int now = q[head++]; if( head == T ) head = 0;        for( int i = last[now]; i; i = e[i].next)            if( e[i].v && e[i].c + d[now] < d[e[i].to] ){                d[e[i].to] = e[i].c + d[now];                from[e[i].to] = i;                if( !inq[e[i].to] ){                    inq[e[i].to] = 1;                    if( d[e[i].to] < d[q[head]] ){                        head--; if( head == -1 ) head = T-1;                        q[head] = e[i].to;                    } else{                        q[tail++] = e[i].to; if( tail == T ) tail = 0;                    }                }            }            inq[now] = 0;    }    return d[T] != inf;}void mcf(){    int x = inf;    for( int i = from[T]; i; i = from[e[i].from] ) x = min(x,e[i].v);    for( int i = from[T]; i; i = from[e[i].from] ){        ans += e[i].c*x;        e[i].v -= x; e[i^1].v += x;    }}int main(){    readIn(n); readIn(m); readIn(k); T = n; S = 0;    for( int i = 1; i <= m; i++ ){        int u, v, w, c;        readIn(u); readIn(v); readIn(w); readIn(c);        insert(u,v,w,c);    } dinic();    printf("%d ", ans);    rebuild();    while(spfa())mcf(); printf("%d", ans);    return 0;}