[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('\n'); 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 }
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……
