Sicily 1034. Forest

1034. Forest

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

In the field of computer science, forest is important and deeply researched , it is a model for many data structures . Now it’s your job here to calculate the depth and width of given forests.

Precisely, a forest here is a directed graph with neither loop nor two edges pointing to the same node. Nodes with no edge pointing to are roots, we define that roots are at level 0 . If there’s an edge points from node A to node B , then node B is called a child of node A , and we define that B is at level (k+1) if and only if A is at level k .

We define the depth of a forest is the maximum level number of all the nodes , the width of a forest is the maximum number of nodes at the same level.

Input

There’re several test cases. For each case, in the first line there are two integer numbers n and m (1≤n≤100, 0≤m≤100, m≤n*n) indicating the number of nodes and edges respectively , then m lines followed , for each line of these m lines there are two integer numbers a and b (1≤a,b≤n)indicating there’s an edge pointing from a to b. Nodes are represented by numbers between 1 and n .n=0 indicates end of input.

Output

For each case output one line of answer , if it’s not a forest , i.e. there’s at least one loop or two edges pointing to the same node, output “INVALID”(without quotation mark), otherwise output the depth and width of the forest, separated by a white space.

Sample Input

1 0
1 1
1 1
3 1
1 3
2 2
1 2
2 1
0 88

Sample Output

0 1
INVALID
1 2
INVALID

Problem Source

ZSUACM Team Member

太久没打代码了导致用队列忘记将元素压入队列……MDZZ

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define rep(i,l,r) for(int i = l; i <= r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(Edge *p = last[x]; p; p = p -> pre)
using namespace std;
const int maxn = 110;
inline int read(){
int ans = 0, f = 1; char c = getchar();
for(; !isdigit(c); c = getchar()) if (c == '-') f = -1;
for(; isdigit(c); c = getchar()) ans = ans * 10 + c - '0';
return ans * f;
}
struct Edge{
Edge *pre; int to;
}edge[10010];
Edge *last[maxn], *pt;
int n, m, inv[maxn], depth[maxn], width[maxn];
bool valid, vis[maxn];
queue <int> q;
inline void addedge(int x,int y){
pt->pre = last[x]; pt->to = y; last[x] = pt++;
}
void bfs(){
while (!q.empty()) q.pop();
rep(i,1,n) if (!inv[i]){
q.push(i); depth[i] = 0; width[0]++; vis[i] = 1;
}
if (q.empty()){
valid = 0; return;
}
while (!q.empty()){
int now = q.front(); q.pop();
travel(now){
if (vis[p->to]){
valid = 0; return;
}
vis[p->to] = 1;
depth[p->to] = depth[now] + 1;
width[depth[p->to]]++;
q.push(p->to);
}
}
rep(i,1,n) if (!vis[i]) valid = 0;
}
void work(){
m = read();
pt = edge; clr(last,0); clr(inv,0); clr(width,0); clr(depth,0); clr(vis,0); valid = 1;
rep(i,1,m){
int a = read(); int b = read();
addedge(a,b);
inv[b]++;
if (inv[b] > 1) valid = 0;
}
if (!valid){
printf("INVALID\n");
return;
}
bfs();
if (!valid){
printf("INVALID\n");
return;
}
int maxdepth = 0; int maxwidth = 0;
rep(i,1,n){
maxdepth = max(maxdepth,depth[i]);
maxwidth = max(maxwidth,width[depth[i]]);
}
printf("%d %d\n",maxdepth,maxwidth);
}
int main(){
n = read();
while (n){
work(); n = read();
}
return 0;
}

View Code