[BZOJ1589][Usaco2008 Dec]Trick or Treat on the Farm 采集糖果
阿新 • • 發佈：2017-09-19
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1 //1有一條邊指向1
3 //2有一條邊指向3
2 //3有一條邊指向2
3
INPUT DETAILS:
Four stalls.
* Stall 1 directs the cow back to stall 1.
* Stall 2 directs the cow to stall 3
* Stall 3 directs the cow to stall 2
* Stall 4 directs the cow to stall 3
2
2
3 縮點之後在DAG上遞推
1589: [Usaco2008 Dec]Trick or Treat on the Farm 采集糖果
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 719 Solved: 408 [Submit][Status][Discuss]Description
每年萬聖節，威斯康星的奶牛們都要打扮一番，出門在農場的N(1≤N≤100000)個牛棚裏轉悠，來采集糖果．她們每走到一個未曾經過的牛棚，就會采集這個棚裏的1顆糖果． 農場不大，所以約翰要想盡法子讓奶牛們得到快樂．他給每一個牛棚設置了一個“後繼牛棚”．牛棚i的後繼牛棚是Xi．他告訴奶牛們，她們到了一個牛棚之後，只要再往後繼牛棚走去，就可以搜集到很多糖果．事實上這是一種有點欺騙意味的手段，來節約他的糖果． 第i只奶牛從牛棚i開始她的旅程．請你計算，每一只奶牛可以采集到多少糖果．Input
第1行輸入N，之後一行一個整數表示牛棚i的後繼牛棚Xi，共N行．Output
共N行，一行一個整數表示一只奶牛可以采集的糖果數量．Sample Input
4 //有四個點1 //1有一條邊指向1
3 //2有一條邊指向3
2 //3有一條邊指向2
3
INPUT DETAILS:
Four stalls.
* Stall 1 directs the cow back to stall 1.
* Stall 2 directs the cow to stall 3
* Stall 3 directs the cow to stall 2
* Stall 4 directs the cow to stall 3
Sample Output
12
2
3 縮點之後在DAG上遞推
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; char buf[10000000], *ptr = buf - 1; inline int readint(){ int n = 0; char ch = *++ptr; while(ch < ‘0‘ || ch > ‘9‘) ch = *++ptr; while(ch <= ‘9‘ && ch >= ‘0‘){ n = (n << 1) + (n << 3) + ch - ‘0‘; ch = *++ptr; } return n; } const int maxn = 200000 + 10; struct Edge{ int to, next; Edge(){} Edge(int _t, int _n): to(_t), next(_n){} }e[maxn]; int fir[maxn] = {0}, cnt = 0; inline void add(int u, int v){ e[++cnt] = Edge(v, fir[u]); fir[u] = cnt; } int n; int dfn[maxn] = {0}, low[maxn], idx = 0; int belong[maxn], bcnt; int sta[maxn], top = 0; bool ins[maxn] = {false}; int f[maxn]; void tarjan(int u){ dfn[u] = low[u] = ++idx; sta[++top] = u; ins[u] = true; for(int v, i = fir[u]; i; i = e[i].next){ v = e[i].to; if(!dfn[v]){ tarjan(v); low[u] = min(low[u], low[v]); } else if(ins[v]) low[u] = min(low[u], dfn[v]); } if(low[u] == dfn[u]){ int now; bcnt++; f[bcnt] = 0; do{ now = sta[top--]; ins[now] = false; belong[now] = bcnt; f[bcnt]++; }while(now != u); } } int ind[maxn] = {0}; int q[maxn], h, t; void tsort(){ h = t = 0; for(int i = n + 1; i <= bcnt; i++) if(!ind[i]) q[t++] = i; int u, v; while(h != t){ u = q[h++]; for(int i = fir[u]; i; i = e[i].next){ v = e[i].to; ind[v]--; if(!ind[v]) q[t++] = v; } } for(int i = t - 1; ~i; i--){ for(int j = fir[q[i]]; j; j = e[j].next) f[q[i]] += f[e[j].to]; } } int main(){ fread(buf, sizeof(char), sizeof(buf), stdin); n = bcnt = readint(); for(int x, i = 1; i <= n; i++){ x = readint(); if(x != i) add(i, x); } for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i); for(int i = 1; i <= n; i++) for(int j = fir[i]; j; j = e[j].next) if(belong[i] != belong[e[j].to]){ add(belong[i], belong[e[j].to]); ind[belong[e[j].to]]++; } tsort(); for(int i = 1; i <= n; i++) printf("%d\n", f[belong[i]]); return 0; }
[BZOJ1589][Usaco2008 Dec]Trick or Treat on the Farm 采集糖果