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POJ 3687 Labeling Balls （拓?fù)渑判颍?/h1>
2019-11-10 21:17:06
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Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:

No two balls share the same label.

The labeling satisfies several constrains like “The ball labeled with a is lighter than the one labeled with b”.

Can you help windy to find a solution?

Input

The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.

Output

For each test case output on a single line the balls’ weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on… If no solution exists, output -1 instead.

Sample Input

54 04 11 14 21 22 14 12 14 13 2

Sample Output

1 2 3 4-1-12 1 3 41 3 2 4

題意

標(biāo)號(hào)為 1~n 的 N 個(gè)球，滿足給定的 M 個(gè)編號(hào)約束關(guān)系，輸出最終滿足關(guān)系的球的標(biāo)號(hào)。

思路

每一個(gè)標(biāo)號(hào)都有可能被其他標(biāo)號(hào)所約束，而對(duì)于這樣的題目我們可以聯(lián)想到拓?fù)渑判颉?/p>

但是題目要求使字典序盡可能的小，于是我們可以逆向建圖，然后從最大的標(biāo)號(hào)開(kāi)始判斷，因?yàn)檫@樣保證了大一點(diǎn)的標(biāo)號(hào)在右邊，于是使得字典序也是最小的了。

AC 代碼

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<vector>#include<queue>#include<set>using namespace std;#define M 210int in[M],arr[M];vector<int>G[M];int main(){ int T; scanf("%d",&T); while(T--) { int n,m; scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) G[i].clear(); memset(in,0,sizeof(in)); for(int i=0; i<m; i++) { int a,b; scanf("%d%d",&a,&b); G[b].push_back(a); //反向建立鄰接表 in[a]++; //點(diǎn)的入度 } int w; for(w=n; w>0; w--) //從最大點(diǎn)開(kāi)始 { int i; for(i=n; i>0; i--) //尋找入度為0的點(diǎn) if(!in[i])break; if(i==0)break; //沒(méi)有找到 arr[i]=w; in[i]=-1; //刪除該點(diǎn) for(int j=0; j<(int)G[i].size(); j++) { int v=G[i][j]; //臨接點(diǎn)的入度-1 if(in[v]>0) in[v]--; } } if(w!=0)

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