zoukankan      html  css  js  c++  java
  • Labeling Balls(拓扑排序wa)

    Labeling Balls
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 12466   Accepted: 3576

    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:

    1. No two balls share the same label.
    2. 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

    5
    
    4 0
    
    4 1
    1 1
    
    4 2
    1 2
    2 1
    
    4 1
    2 1
    
    4 1
    3 2
    

    Sample Output

    1 2 3 4
    -1
    -1
    2 1 3 4
    1 3 2 4
    题解:反向建图,却是wa。。。先放着吧
    wa代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<queue>
     4 #define mem(x,y) memset(x,y,sizeof(x)) 
     5 using namespace std;
     6 const int MAXN=210;
     7 const int MAXM=40010<<1;
     8 int head[MAXM];
     9 int que[MAXN],ans[MAXN];
    10 struct Edge{
    11     int from,to,next;
    12 }; 
    13 Edge edg[MAXM];
    14 int edgnum;
    15 int N;
    16 void add(int u,int v){
    17     Edge E={u,v,head[u]};
    18     edg[edgnum]=E;
    19     head[u]=edgnum++;
    20 } 
    21 void topu(){
    22     priority_queue<int>dl;
    23     int top=1;
    24     for(int i=1;i<=N;i++)
    25         if(!que[i])dl.push(i);
    26     while(!dl.empty()){
    27         int k=dl.top();
    28         dl.pop();
    29         ans[top++]=k;
    30         que[k]=-1;
    31         for(int i=head[k];i!=-1;i=edg[i].next){
    32             int v=edg[i].to;
    33             que[v]--;
    34             if(!que[v])dl.push(v);
    35         }
    36     }
    37     if(top!=N+1)puts("-1");
    38     else{
    39     for(int i=N;i>=1;i--){
    40         if(i!=N)printf(" ");
    41         printf("%d",ans[i]);
    42     }
    43     puts("");
    44     }
    45 }
    46 int main(){
    47     int T,M,a,b;
    48     scanf("%d",&T);
    49     while(T--){
    50         scanf("%d%d",&N,&M);
    51         for(int i=1;i<=N;i++)ans[i]=i;
    52         mem(que,0);
    53         mem(head,-1);
    54         edgnum=0;
    55         while(M--){
    56             scanf("%d%d",&a,&b);
    57             add(b,a);//
    58             que[a]++;//反向建图 
    59         }
    60         topu();
    61     }
    62     return 0;
    63 }
  • 相关阅读:
    2017.10.16 java中getAttribute和getParameter的区别
    2017.10.15 解析Java中抽象类和接口的区别
    2017.10.14 Java的流程控制语句switch&&随机点名器
    2017.10.13 Java中引用类型变量的创建及使用&循环高级写法
    2017.10.12 Java的计数器的开发
    2017.10.11 ”学生身体体质信息管理系统“的开发
    2017.10.10 java中的继承与多态(重载与重写的区别)
    一次redis悲观锁 实现 微信jssdk token缓存
    HashSet原理 与 linkedHashSet
    jdk并发容器整理(yet)
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4822313.html
Copyright © 2011-2022 走看看