poj3687Labeling Balls【反向拓扑排序 模板】

Labeling Balls

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12994 Accepted: 3748

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 toN 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 withb".

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 nextM line each contain two integers a and b indicating the ball labeled witha 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 labelN. 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

折腾了将近两个小时总算在discuess的只言片语中知道自己错在哪里了，怎么说也是没参考别人代码A的，但是思维严谨性啊……自己写测试数据！自己好好读题！

1.题中有一句特别坑爹的话：For each test case output on a single line the balls' weights from label 1 to labelN. 输出的是这个数字所在的位置 ，不是这个位置上的数字

2.反向输出！这个写几组数据就能看出来吧。但是为什么呢？因为如果只是限制优先级，小的数先输出，有可能得不到正解，反向输出，大的放到后面就解决了这个问题

/****************poj36872016.2.3916K79MSG++1939B****************/#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include<vector>using namespace std;const int maxn=300;int tmp[400];bool mark[205][205];int head[maxn],ip,indegree[maxn];int t,n,m,seq[maxn];struct cmp1{    bool operator ()(int &a,int &b){        return a<b;//最大值优先    }};struct note{    int v,next;} edge[maxn*maxn];void init(){    memset(head,-1,sizeof(head));    ip=0;    memset(indegree,0,sizeof(indegree));    memset(mark,0,sizeof(mark));}void addedge(int u,int v){    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;}int topo()///拓扑，可做模板{    priority_queue<int,vector<int>,cmp1>que1;    int indeg[maxn];    for(int i=1; i<=n; i++)    {        indeg[i]=indegree[i];        if(indeg[i]==0)            que1.push(i);    }    int k=n;    bool res=false;    while(!que1.empty())    {        if(que1.size()!=1)res=true;        int u=que1.top();        que1.pop();        seq[k--]=u;        for(int i=head[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            indeg[v]--;            if(indeg[v]==0)                que1.push(v);        }    }    if(k>0)return -1;///存在有向环，总之不能进行拓扑排序    if(res)return 0;///可以进行拓扑排序，并且只有唯一一种方式，seq数组即是排序完好的序列    return 1;///可以进行拓扑排序，有多种情况，seq数组是其中一种序列}int main(){    、、freopen("cin.txt","r",stdin);    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        init();        while(m--)        {            int a,b;            scanf("%d%d",&a,&b);            if(mark[a][b]) continue;            addedge(b,a);            indegree[a]++;            mark[a][b]=1;        }        if(topo()==-1) printf("-1\n");        else        {            for(int i=1;i<=n;i++) tmp[seq[i]]=i;            for(int i=1;i<n;i++) printf("%d ",tmp[i]);            printf("%d\n",tmp[n]);        }    }    return 0;}