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Icontofig 's Blog // 我们的征程是星辰大海!Away From OI,Come to ICPC(查看友链请点About Me)

 2019-07-07 21:56:00 |  0 Comments  |  topsort DP

洛谷P1137 旅行计划

题解

别问我为什么写这么水的题解,问就是为了那门水的不行的专业选修课。

很简单,这个图肯定是个DAG(请不要问为什么,自己看一下题目描述。。。)

然后我们在这个DAG上按拓扑序来进行DP就可以了,因为入度为0的点一定最多游览一个,然后不断类似SPFA最长路一样的进行DP,就可以求出到某个城市最多可以游览的城市数目了。

代码

#include <bits/stdc++.h>
using namespace std;
int get_num(){
    int num = 0;
    char c;
    bool flag = false;
    while((c = getchar()) == ' ' || c == '\r' || c == '\n');
    if(c == '-')
        flag = true;
    else num = c - '0';
    while(isdigit(c = getchar()))
        num = (num<<3) + (num<<1) + c - '0';
    return (flag ? -1 : 1) * num;
}
int n,m;
int x,y;
const int maxn = 1e5+5;
vector<int>g[maxn];
int f[maxn],d[maxn];
int inq[maxn];
int main(){
    memset(d,0,sizeof(d));
    n = get_num();
    m = get_num();
    for(int i = 1;i <= m;++i){
        x = get_num();
        y = get_num();
        g[x].push_back(y);
        d[y]++;
    }
    queue<int>q;
    for(int i = 1;i <= n;++i){
        if(!d[i])
            q.push(i),inq[i] = 1;
        f[i] = 1;
    }
    while(!q.empty()){
        int x = q.front();
        q.pop();
 2019-02-26 18:14:48 |  0 Comments  |  topsort

HDU - 4948 Kingdom 拓扑排序

Description

Teacher Mai has a kingdom consisting of n cities. He has planned the transportation of the kingdom. Every pair of cities has exactly a one-way road.

He wants develop this kingdom from one city to one city.

Teacher Mai now is considering developing the city w. And he hopes that for every city u he has developed, there is a one-way road from u to w, or there are two one-way roads from u to v, and from v to w, where city v has been developed before.

He gives you the map of the kingdom. Hope you can give a proper order to develop this kingdom.

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains an integer n (1<=n<=500).

The following are n lines, the i-th line contains a string consisting of n characters. If the j-th characters is 1, there is a one-way road from city i to city j.

Cities are labelled from 1.

Output

If there is no solution just output "-1". Otherwise output n integers representing the order to develop this kingdom.

Sampl

 2019-02-07 01:05:31 |  0 Comments  |  2-SAT tarjan topsort

BUPT校内训练&CodeChef - VRTXCOVR-Vertex Cover 2-SAT

Description

You are given an undirected graph G = (V, E) containing N nodes and M edges. The nodes are numbered from 1 to N. A subset C of V is a vertex cover if for every edge (u, v) ∈ E, at least one of u and v belong to C. Note that C = V is always a vertex cover.
Consider a partition of V into two sets A and B. It is said to be a valid partition, if the following two conditions are satisfied: A should be a vertex cover. And for each i such that 1 ≤ i ≤ n/2, nodes 2i and 2i - 1 don't belong to the same set (i.e. one belongs to set A and the other to set B).
Determine if a valid partition exists. If it exists, provide an example of one valid partition.

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains two space-separated integers N and M denoting the number of nodes and number of edges in the graph respectively.
Each of the following M lines contains two space-se

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