首页 > 代码库 > 费用流做题记录

费用流做题记录

BZOJ1221:http://www.lydsy.com/JudgeOnline/problem.php?id=1221

  trick:将每天用完的,和要用的分来开处理,避免花费的重叠计算,也就是拆点

技术分享
#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair<int,int>#define MP make_pairtypedef long long LL;const long long INF = 1e18+1LL;const double Pi = acos(-1.0);const int N = 4e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e7;int na[N],ans,ans1,S,T,head[N],cnt = 1,n,dis[N],q[N],inq[N],from[N],f,fA,fB;struct {    int from,to,next,c,v;}e[N * 2];void ins(int u,int v,int w,int c) {    cnt++;    e[cnt].from = u;    e[cnt].to = v;    e[cnt].v = w;    e[cnt].c = c;    e[cnt].next = head[u];    head[u] = cnt;}void insert(int u,int v,int w,int c) {    ins(u,v,w,c);ins(v,u,0,-c);}int spfa() {    for(int i = 0; i <= T; ++i) dis[i] = inf;    int t = 0,w = 1;    dis[S] = q[S] = S;    inq[S] = 1;    while(t!=w) {        int now = q[t++];        if(t == 200001) t = 0;        for(int i = head[now]; i; i = e[i].next) {            if(e[i].v&&dis[e[i].to] > dis[now] + e[i].c)            {                from[e[i].to] = i;                dis[e[i].to] = dis[now] + e[i].c;                if(!inq[e[i].to]) {                    inq[e[i].to] = 1;                    q[w++] = e[i].to;                    if(w == 200001) w = 0;                }            }        }        inq[now] = 0;    }    if(dis[T] >= inf) return 0;    return 1;}void mcf() {    int i = from[T],x=inf;    while(i) {        x = min(e[i].v,x);        i = from[e[i].from];    }    i = from[T];    ans1 = x;    while(i) {        e[i].v -= x;        e[i^1].v += x;        ans += e[i].c*x;        i = from[e[i].from];    }}int a,b;int main(){    cnt = 1;    memset(head,0,sizeof(head));    scanf("%d%d%d%d%d%d",&n,&a,&b,&f,&fA,&fB);    S = 0,T = 3*n;    for(int i = 1; i <= n; ++i){        scanf("%d",&na[i]);        insert(S,i,inf,0);        insert(S,i+n,inf,f);        insert(i+n,T,na[i],0);        if(i + 1 <= n)            insert(i,i+1,inf,0);        if(i + a + 1 <= n)            insert(i,n+i+a+1,inf,fA);        if(i + b + 1 <= n)            insert(i,n+i+b+1,inf,fB);    }    ans = 0,ans1 = 0;    while(spfa()) mcf();    printf("%d\n",ans);    return 0;}
BZOJ1221

 

  

费用流做题记录