编程练习8(多源最短路)
2016-11-28 23:58
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A:Stockbroker Grapevine
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %lluDescription
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 5
2 1 2 3 6
2 1 2 2 2
5
3 4 4 2 8 5 3
1 5 8
4 1 6 4 10 2 7 5 2
0
2 2 5 1 5
0
Sample Output
3 23 10
我的代码
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> using namespace std; typedef long long ll; const double PI = acos(-1.0); const double eps = 1e-6; const int INF = 1000000000; const int maxn = 100; int P,n,m; int i,j; int dis[105][105]; int a,b; void init(){ for (i = 0 ; i < 105 ; i++) { for (j = 0; j < 105; j++) { if (j == i) { dis[i][j] = 0; }else{ dis[i][j] = INF; } } } } struct result{ int start; int cost; }; void floyd(){ int i,j,k; for (k = 1; k <= P; k++) { for (i = 1; i <= P; i++) { for (j = 1; j <=P; j++) { if (dis[i][j] > dis[i][k]+dis[k][j]) { dis[i][j] = dis[i][k]+dis[k][j]; } } } } } int main() { freopen("input.txt", "r", stdin); while(scanf("%d",&P)!=EOF && P) { init(); //input for (i = 1; i <= P ; i++) { cin >> n; for (j = 1; j <=n; j++) { cin>>a>>b; dis[i][a] = b; } } //floyd floyd(); //找到答案 result re,temp; re.cost = INF; for (i = 1; i <= P; i++) { temp.cost = 0; for (j = 1; j <= P; j++) { if (dis[i][j] > temp.cost) { temp.cost = dis[i][j]; temp.start = i; } } if (temp.cost < re.cost) { re = temp; } } if (re.cost == INF) { cout <<"disjoint" << endl; }else{ cout << re.start <<" " << re.cost << endl; } } return 0; }
B:一个人的旅行(HDU2066)
Description
虽然草儿是个路痴(就是在杭电待了一年多,居然还会在校园里迷路的人,汗~),但是草儿仍然很喜欢旅行,因为在旅途中 会遇见很多人(白马王子,^0^),很多事,还能丰富自己的阅历,还可以看美丽的风景……草儿想去很多地方,她想要去东京铁塔看夜景,去威尼斯看电影,去阳明山上看海芋,去纽约纯粹看雪景,去巴黎喝咖啡写信,去北京探望孟姜女……眼看寒假就快到了,这么一大段时间,可不能浪费啊,一定要给自己好好的放个假,可是也不能荒废了训练啊,所以草儿决定在要在最短的时间去一个自己想去的地方!因为草儿的家在一个小镇上,没有火车经过,所以她只能去邻近的城市坐火车(好可怜啊~)。Input
输入数据有多组,每组的第一行是三个整数T,S和D,表示有T条路,和草儿家相邻的城市的有S个,草儿想去的地方有D个;接着有T行,每行有三个整数a,b,time,表示a,b城市之间的车程是time小时;(1=<(a,b)<=1000;a,b 之间可能有多条路)
接着的第T+1行有S个数,表示和草儿家相连的城市;
接着的第T+2行有D个数,表示草儿想去地方。
Output
输出草儿能去某个喜欢的城市的最短时间。Sample Input
6 2 31 3 5
1 4 7
2 8 12
3 8 4
4 9 12
9 10 2
1 2
8 9 10
Sample Output
9我的代码
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> using namespace std; typedef long long ll; const double PI = acos(-1.0); const double eps = 1e-6; const int INF = 1000000000; const int maxn = 100; int T,S,D; int i,j; int a,b,c; int edge[1005][1005]; bool visited[1005]; int dis[1005]; int maxnum; struct result{ int city; int cost; }; void init(){ for (i = 0; i < 1005; i++) { visited[i] = false; } for (i = 0; i < 1005; i++) { dis[i] = INF; } for (i = 0; i < 1005; i++) { for (j = 0; j < 1005; j++) { if (i == j) { edge[i][j] = 0; }else{ edge[i][j] = INF; } } } } void dijstra(){ visited[0] = true; dis[0] = 0; int miny = 0,temp; int count = 0; //要考虑孤立点 while(1){ //changed = false; count++; if (count > 1005) { break; } temp = INF; for (i = 0; i <= maxnum+1; i++) { if (!visited[i] && dis[i] < temp) { // changed = true; miny = i; temp = dis[i]; } } if (miny == maxnum+1) { break; } visited[miny] = true; for (i = 1; i <= maxnum+1; i++) { if (!visited[i] && dis[i] > dis[miny]+edge[miny][i]) { dis[i] = dis[miny]+edge[miny][i]; } } } } int main() { //freopen("input.txt", "r", stdin); while(scanf("%d%d%d",&T,&S,&D)!=EOF) { init(); maxnum = 0; for (i = 0; i < T; i++) { scanf("%d%d%d",&a,&b,&c); //无向图啊!两条边都要赋值啊!!! if (c < edge[a][b]) { edge[a][b]= edge[b][a]=c; } if (a > maxnum) { maxnum = a; } if (b > maxnum) { maxnum = b; } } int num; for (i = 0; i < S; i++) { scanf("%d",&num); edge[0][num] = 0; } int des; for (i = 0; i < D; i++) { scanf("%d",&des); edge[des][maxnum+1] = 0; } dijstra(); printf("%d\n",dis[maxnum+1]); } return 0; }
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