[floyd]poj1125 Stockbroker Grapevine

Stockbroker Grapevine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 34129 Accepted: 18867

Description

Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to spread the rumours in the fastest possible way. 

Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.

Input

Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, and the time taken for them to pass the message to each person. The format of each stockbroker line is as follows: The line starts with the number of contacts (n), followed by n pairs of integers, one pair for each contact. Each pair lists first a number referring to the contact (e.g. a '1' means person number one in the set), followed by the time in minutes taken to pass a message to that person. There are no special punctuation symbols or spacing rules. 

Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people. 

Output

For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes. 
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.

Sample Input

32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50

Sample Output

3 23 10

Source

Southern African 2001

题意：

For maximum effect, you have to spread the rumours in the fastest possible way. 

为了获得最大的效果，你必须以最快的方式传播谣言。

 This duration is measured as the time needed for the last person to receive the information.

此持续时间为最后一个接收信息的人所需的时间。

思路：

多源最短路，有向图，跟 poj2253 frogger 类型很像，同样直接floyd即可；

本题并不是传统意义上的最短路，poj2253 frogger 提到了一个概念“minimax”(极小极大值，即极大值中最小的)；

你要求每个stockbroker传播所有能传播的人之后所需要的最长时间，然后在这些最长时间中求出最短的时间，最短时间对应的stockbroker就是我们要找的人

代码：

#include <iostream>#include <stdio.h>#include <string.h>#define inf 1<<29//求的是最快的using namespace std;const int N = 110;int a[N][N];void floyd(int n){    int i, j, k;    for(k = 1; k <= n; k++)    for(i = 1; i <= n; i++)    for(j = 1; j <= n; j++)        if(a[i][j] > a[i][k] + a[k][j])            a[i][j] = a[i][k] + a[k][j];}int main(){    int n, m;    int from, to, value;    int i, j;    while(scanf("%d", &n) && n)    {        //预处理        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)                a[i][j] = inf;            a[i][i] = 0;        }        //录入数据        for(from = 1; from <= n; from++)        {            scanf("%d", &m);            while(m-->0)            {                scanf("%d%d", &to, &value);                if(a[from][to] > value)                    a[from][to] = value;            }        }        //floyd        floyd(n);        int t, Max, Min = inf;        for(i = 1; i <= n; i++)        {            Max = 0;            for(j = 1; j <= n; j++)                Max = max(Max, a[i][j]);            if(Min > Max)            {                Min = Max;                t = i;            }        }        if(Min > Max)        {            printf("disjoint\n");            continue;        }        printf("%d %d\n", t, Min);    }    return 0;}/*32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50----------3 23 10disjoint*/

反思：

解释一下为什么floyd的k层循环要放在最外层：

我们不要忘记floyd的算法的本质是动态规划，k是阶段，i,j是状态；

如果把k层循环放里面，就会过早的把i到j的最短路径确定下来了，如果后面存在更短的路径时，就不会再更新了；

举个例子：

A->B, 如果k层循环放里面，那么A->B的最短路径就永远固定在9了，显然其最短路径应该是6

参考博客：http://www.cnblogs.com/twjcnblog/archive/2011/09/07/2170306.html

Stockbroker Grapevine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 34129 Accepted: 18867

Description

Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to spread the rumours in the fastest possible way. 

Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.

Input

Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, and the time taken for them to pass the message to each person. The format of each stockbroker line is as follows: The line starts with the number of contacts (n), followed by n pairs of integers, one pair for each contact. Each pair lists first a number referring to the contact (e.g. a '1' means person number one in the set), followed by the time in minutes taken to pass a message to that person. There are no special punctuation symbols or spacing rules. 

Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people. 

Output

For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes. 
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.

Sample Input

32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50

Sample Output

3 23 10

Source

Southern African 2001

题意：

For maximum effect, you have to spread the rumours in the fastest possible way. 

为了获得最大的效果，你必须以最快的方式传播谣言。

 This duration is measured as the time needed for the last person to receive the information.

此持续时间为最后一个接收信息的人所需的时间。

思路：

多源最短路，有向图，跟 poj2253 frogger 类型很像，同样直接floyd即可；

本题并不是传统意义上的最短路，poj2253 frogger 提到了一个概念“minimax”(极小极大值，即极大值中最小的)；

你要求每个stockbroker传播所有能传播的人之后所需要的最长时间，然后在这些最长时间中求出最短的时间，最短时间对应的stockbroker就是我们要找的人

代码：

#include <iostream>#include <stdio.h>#include <string.h>#define inf 1<<29/*求的是最快的，*/using namespace std;const int N = 110;int a[N][N];void floyd(int n){    int i, j, k;    for(k = 1; k <= n; k++)    for(i = 1; i <= n; i++)    for(j = 1; j <= n; j++)        if(a[i][j] > a[i][k] + a[k][j])            a[i][j] = a[i][k] + a[k][j];}int main(){    int n, m;    int from, to, value;    int i, j;    while(scanf("%d", &n) && n)    {        //预处理        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)                a[i][j] = inf;            a[i][i] = 0;        }        //录入数据        for(from = 1; from <= n; from++)        {            scanf("%d", &m);            while(m-->0)            {                scanf("%d%d", &to, &value);                if(a[from][to] > value)                    a[from][to] = value;            }        }        //floyd        floyd(n);        int t, Max, Min = inf;        for(i = 1; i <= n; i++)        {            Max = 0;            for(j = 1; j <= n; j++)                Max = max(Max, a[i][j]);            if(Min > Max)            {                Min = Max;                t = i;            }        }        if(Min > Max)        {            printf("disjoint\n");            continue;        }        printf("%d %d\n", t, Min);    }    return 0;}/*32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50----------3 23 10disjoint*/

反思：

解释一下为什么floyd的k层循环要放在最外层：

我们不要忘记floyd的算法的本质是动态规划，k是阶段，i,j是状态；

如果把k层循环放里面，就会过早的把i到j的最短路径确定下来了，如果后面存在更短的路径时，就不会再更新了；

举个例子：

A->B, 如果k层循环放里面，那么A->B的最短路径就永远固定在9了，显然其最短路径应该是6

参考博客：http://www.cnblogs.com/twjcnblog/archive/2011/09/07/2170306.html