Algorithm

SPFA:Til the Cows Come Home

October 27, 2012 Graph Theory No comments

D – Til the Cows Come Home
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Description
Bessie is out in[……]

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BellmanFord: Til the cows come home

October 27, 2012 Graph Theory No comments

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 20995 Accepted: 6994

Description
Bessie is out in the field and wants to get b[……]

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Kruskal: Matrix

October 27, 2012 Graph Theory No comments

Problem Description
Machines have once again attacked the kingdom of Xions. The kingdom of Xions has N cities and N-1 bidirectional roads. The road n[……]

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SPFA: Power Transmission

October 27, 2012 Graph Theory No comments

Problem Description

The project West-East power transmission is famous around the world. It transmits the electricity from western areas to east Ch[……]

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Dijkstra: Till the Cows Come Home

October 27, 2012 Graph Theory No comments

D – Til the Cows Come Home
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Description
Bessie is out in[……]

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Kruscal: Agri-Net

October 27, 2012 Graph Theory No comments

B – Agri-Net
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Description
Farmer John has been elected ma[……]

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Prim: Agri-Net

October 27, 2012 Graph Theory No comments

B – Agri-Net
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Description
Farmer John has been elected ma[……]

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October 3, 2012 Graph Theory No comments

eg

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#define MAX 10000
 
using namespace std;
 
int cnt;//0
int lastshow[MAX];//i,-1
int sum;
int vis[MAX];
 
struct edge{   //a->b
    int to;    //b
    int pro;  //a
    int weight;//
}e[MAX];
 
void insert(){
    int a, b, weight;
    scanf("%d %d %d", &a, &b, &weight);
    e[cnt].to = b;
    e[cnt].pro = lastshow[a]; //a
    e[cnt].weight = weight;
    lastshow[a] = cnt ++; //lastshow
}
 
int dfs(int a){
    if(vis[a])
        return 0;
//    cout << endl << a << endl;
    vis[a] = 1;
    int ret = 0;
    int id = lastshow[a];
    while(id != -1){
        ret += e[id].weight;
//        cout << endl << e[id].weight << endl;
        ret += dfs(e[id].to);
        id = e[id].pro;
    }
    return ret;
}
 
int main(){
    int n, c, srch;
    int i, j, k;
    while(cin >> n >> c){
        cnt = 0;
        sum = 0;
        memset(vis, 0, sizeof(vis));
        for(i = 0; i < MAX; i ++){
            lastshow[i] = -1;
        }
        for(i = 0; i < c; i ++){
            insert();
        }
        //i
        while(~scanf("%d", &srch)){
            if(-1 == srch)
                break;
            int id = lastshow[srch];
            cout << "" << srch << ":"<<endl;
            while(id != -1){
                cout << e[id].to << " " << e[id].weight << endl;
                id = e[id].pro;
            }
        }
        cout << "" << dfs(0) << endl;
    }
    return 0;
}

Color A Tree

October 3, 2012 Graph Theory No comments

Description
Bob is very interested in the data structure of a tree. A tree is a directed graph in which a special node is singled out, called the “ro[……]

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+Toy Storage

October 3, 2012 Computational Geometry No comments

Description
Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangul[……]

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: Intersection

October 3, 2012 Computational Geometry No comments

IntersectionTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Description
You are to write a program that h[……]

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