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  • [BZOJ 3218]a + b Problem

    又是一道主席树优化网络流的好题

    按约大爷的教导,源点为白,汇点为黑,搞成最小割

    发现暴力连边要爆炸,但是要连的点在线段树中都构成了一个区间,果断主席树优化之

    为什么不用一般线段树?

    因为要满足 j<i ,这里的可持久化并不是为了查询过去的值,而是为了保留过去的值不与后来弄混~

    如果有两个点的 a[i] 相同在线段树里怎么搞?

    很简单,从 a[i] 向 a[j] 连一条 inf 的流即可

    不过——为什么题目名字那么坑啊啊啊啊啊啊?!!!!!

        这种题是不是非要来卡空间不然不痛快是吗?!!!!!

      1 #include <cstdio>
      2 #include <cstring>
      3 const int sizeOfSegment=500005;
      4 const int sizeOfEdge=1000001;
      5 const int sizeOfPoint=500005;
      6 const int sizeOfCell=5005;
      7 const int inf=1000000000;
      8 
      9 inline int min(int, int);
     10 inline int getint();
     11 inline void putint(int);
     12 
     13 int S, T;
     14 int V;
     15 int n;
     16 int a[sizeOfCell], b[sizeOfCell], w[sizeOfCell], l[sizeOfCell], r[sizeOfCell], p[sizeOfCell];
     17 
     18 struct edge {int point, flow; edge * next, * pair;};
     19 edge memory_edge[sizeOfEdge], * port_edge=memory_edge;
     20 edge * e[sizeOfPoint];
     21 inline edge * newedge(int, int, edge * );
     22 inline void link(int, int, int);
     23 int h[sizeOfPoint], gap[sizeOfPoint];
     24 inline bool bfs();
     25 inline int isap();
     26 
     27 struct seg {int p; seg * l, * r;};
     28 seg memory_seg[sizeOfSegment], * port_seg=memory_seg;
     29 seg * t;
     30 inline seg * newseg(seg * =NULL);
     31 seg * insert(seg * , int, int, int, int);
     32 void query(seg * , int, int, int, int, int);
     33 
     34 int main()
     35 {
     36     int ans=0;
     37 
     38     n=getint();
     39     S=0; T=n+1; V=n+1;
     40     for (int i=1;i<=n;i++)
     41     {
     42         a[i]=getint(), b[i]=getint(), w[i]=getint(), l[i]=getint(), r[i]=getint(), p[i]=getint();
     43         ans+=b[i]+w[i];
     44         link(S, i, b[i]); link(i, T, w[i]);
     45         link(i, ++V, p[i]);
     46         query(t, 0, inf, l[i], r[i], V);
     47         t=insert(t, 0, inf, a[i], i);
     48     }
     49 
     50     ans-=isap();
     51     putint(ans);
     52 
     53     return 0;
     54 }
     55 inline int min(int x, int y)
     56 {
     57     return x<y?x:y;
     58 }
     59 inline int getint()
     60 {
     61     register int num=0;
     62     register char ch;
     63     do ch=getchar(); while (ch<'0' || ch>'9');
     64     do num=num*10+ch-'0', ch=getchar(); while (ch>='0' && ch<='9');
     65     return num;
     66 }
     67 inline void putint(int num)
     68 {
     69     char stack[11];
     70     register int top=0;
     71     if (num==0) stack[top=1]='0';
     72     for ( ;num;num/=10) stack[++top]=num%10+'0';
     73     for ( ;top;top--) putchar(stack[top]);
     74     putchar('
    ');
     75 }
     76 inline edge * newedge(int point, int flow, edge * next)
     77 {
     78     edge * ret=port_edge++;
     79     ret->point=point; ret->flow=flow; ret->next=next;
     80     return ret;
     81 }
     82 inline void link(int u, int v, int f)
     83 {
     84     e[u]=newedge(v, f, e[u]); e[v]=newedge(u, 0, e[v]);
     85     e[u]->pair=e[v]; e[v]->pair=e[u];
     86 }
     87 inline bool bfs()
     88 {
     89     static int q[sizeOfPoint];
     90     static int l, r;
     91     memset(h, 0xFF, sizeof(h)); h[T]=0;
     92     l=r=0;
     93     for (q[r++]=T;l<r;l++)
     94     {
     95         int u=q[l];
     96         ++gap[h[u]];
     97         for (edge * i=e[u];i;i=i->next) if (h[i->point]==-1)
     98         {
     99             h[i->point]=h[u]+1;
    100             q[r++]=i->point;
    101         }
    102     }
    103     return h[S]>-1;
    104 }
    105 inline int isap()
    106 {
    107     static edge * t[sizeOfPoint], * p[sizeOfPoint];
    108     static int aug[sizeOfPoint];
    109     int flow=0, hmin=0;
    110 
    111     if (!bfs()) return 0;
    112 
    113     memcpy(t, e, sizeof(e)); memset(p, 0, sizeof(p));
    114     aug[S]=inf;
    115     for (int u=S;h[S]<V; )
    116     {
    117         if (u==T)
    118         {
    119             flow+=aug[T];
    120             for (edge * i=p[T];i;i=p[i->point])
    121                 aug[i->point]-=aug[T], i->pair->flow-=aug[T], i->flow+=aug[T];
    122             for (edge * i=p[T];i;i=p[i->point]) if (aug[i->point])
    123             {
    124                 u=i->point;
    125                 break;
    126             }
    127         }
    128 
    129         edge *& i=t[u];
    130         for ( ;i && (!i->flow || h[i->point]+1!=h[u]);i=i->next);
    131         if (i)
    132         {
    133             aug[i->point]=min(aug[u], i->flow); p[i->point]=i->pair;
    134             u=i->point;
    135         }
    136         else
    137         {
    138             if (!--gap[h[u]]) break;
    139             hmin=V;
    140             for (edge * j=e[u];j;j=j->next) if (j->flow && h[j->point]+1<hmin)
    141             {
    142                 hmin=h[j->point]+1;
    143                 t[u]=j;
    144             }
    145             ++gap[h[u]=hmin];
    146             u=u==S?S:p[u]->point;
    147         }
    148     }
    149 
    150     return flow;
    151 }
    152 inline seg * newseg(seg * t)
    153 {
    154     seg * newt=port_seg++;
    155     newt->p=++V; newt->l=t?t->l:NULL; newt->r=t?t->r:NULL;
    156     return newt;
    157 }
    158 seg * insert(seg * t, int l, int r, int p, int v)
    159 {
    160     seg * newt=newseg(t);
    161     if (l==r)
    162     {
    163         if (t) link(newt->p, t->p, inf);
    164         link(newt->p, v, inf);
    165     }
    166     else
    167     {
    168         int m=(l+r)>>1;
    169         if (p<=m)
    170         {
    171             newt->l=insert(newt->l, l, m, p, v), link(newt->p, newt->l->p, inf);
    172             if (newt->r) link(newt->p, newt->r->p, inf);
    173         }
    174         else
    175         {
    176             newt->r=insert(newt->r, m+1, r, p, v), link(newt->p, newt->r->p, inf);
    177             if (newt->l) link(newt->p, newt->l->p, inf);
    178         }
    179     }
    180 
    181     return newt;
    182 }
    183 void query(seg * t, int l, int r, int ql, int qr, int v)
    184 {
    185     if (!t) return ;
    186     if (l==ql && r==qr) link(v, t->p, inf);
    187     else
    188     {
    189         int m=(l+r)>>1;
    190         if (qr<=m) query(t->l, l, m, ql, qr, v);
    191         else if (ql>=m+1) query(t->r, m+1, r, ql, qr, v);
    192         else query(t->l, l, m, ql, m, v), query(t->r, m+1, r, m+1, qr, v);
    193     }
    194 }
    改数组改到想去死

    尝试着用了用 Data Display Debugger 觉得是很强大但是总觉得也很不顺手?

    不过反正有了这个我是不想再去用 gdb 了……

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  • 原文地址:https://www.cnblogs.com/dyllalala/p/4145272.html
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