多源最短路训练题解(floyd模板+ 无向图+ 有向图+10多道类型题解)
还有16天就要比省赛了，复习一波最短路首先是floyd的应用
需要注明的一点第一个K是松弛了k个城市
练习题解:
HDU1596
HDU2112
HDU1874
HDU1869
HDU2066
HDU2094
HDU2544
稍加复杂：
HDU1217 顺练习map离散 难度1.5
HDU1245 处理起点，终点 难度2.5
HDU1535 2次迪杰斯特拉 难度2
HDU2170 难度x
HDU3631 难度x
HDU4284 BFS+floyd 难度2.5
加深对k的理解：
HDU1599（最小环）
暴富的小灰
时间限制: 1 Sec 内存限制: 128 MB
提交: 30 解决: 12
[提交] [状态] [讨论版] [命题人:外部导入]
题目描述
小灰怎么可能暴富呢！！！最近上映了一部电影，叫做“西虹市首富”，小灰看完电影做了个梦，梦到自己被要求在一个月内花光十个亿的鼠币，它决定带着小白去旅行，从X城一路旅行到Y城，小灰想知道怎么走才能花最多的钱，你能帮助小灰吗？
输入
第一行三个正整数N，X，Y，分别表示城市的个数，旅行的起点，旅行的终点；第二行N个正整数K，表示每个城市旅行的花费；接下来的N-1行，每行有两个正整数U，V，表示能从U城市到达V城市。
输出
输出小灰最多能花费多少钱，如果从X城市不能到达Y城市，则输出-1
样例输入
6 1 6
7 3 4 2 8 5
1 2
2 3
3 6
4 2
5 2
样例输出
19提示
1 <= N <= 150
1 <= K <= 10000
floyd裸题，花费弄成负的转化成最短路
# include
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# include
Travel Cost
时间限制: 1 Sec 内存限制: 128 MB
提交: 38 解决: 21
[提交] [状态] [讨论版] [命题人:*Administrator]
题目描述
AC country is a famous and beautiful place. There are N cities in the country, numbered as 1,2,3...N。The first city is the capital of AC, so it is the greatest and best place all over the country. Many travelers often get a travel in AC. ACMer and his girl friends like the fine view and beautiful scenery in AC, so they plan to get a travel in AC recently. AC and his N-1 girl friends go to the AC by plane land at the capital airport. They wants to get a travel to every city of AC but independently. in other words, the first girl friend goes to the first city, the second girl friend goes to the second city，the rest can be done in the same manner.
In AC country, there maybe at most one road between every two different cities. One girl will spend some money if she passes one road correspondingly. ACMer must offer all the travel cost of all his girl friends. He wants to known what is the minimum total cost if evry girl gets her travel by the most economic strategy ?
输入
There are many test case waiting for your program code!
In every case, the first line of the input will be N (1<=N<=100), the number of cites.
The rest of the input defines an N*N adjacency matrix. Each of its entries will be either an integer or the character x. The value of MAT(i,j) indicates the expense of sending a travel directly from city i to city j. A value of x for MAT(i,j) indicates that there is no directly road from city to city j.
Note that for a city to itself does not require any cost, so MAT(i,i) = 0 for all integer i in range(1,N). Also, you may assume that the road is undirected, so that MAT(i,j) = MAT(j,i). Thus the imput data only supply the entries on the lower triangular portion of adjacency matrix strictly.
The input of your program will be the lower triangular section of matrix . That is, the second line of input will contain only, MAT(2,1). The next line will contain two entries, MAT(3,1) and MAT(3,2), and so on. All MAT(i,j) will not greater than 1000.
输出
for every case, output the minimum total cost of all ACMer's all girl friends, one per line.
样例输入
5
40
30 6
99 24 36
25 x x 25
样例输出
141
裸的最短路
# include
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find the safest roadTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18643 Accepted Submission(s): 6364 Problem Description XX星球有很多城市，每个城市之间有一条或多条飞行通道，但是并不是所有的路都是很安全的，每一条路有一个安全系数s,s是在 0 和 1 间的实数(包括0，1)，一条从u 到 v 的通道P 的安全度为Safe(P) = s(e1)*s(e2)…*s(ek) e1,e2,ek是P 上的边 ，现在8600 想出去旅游，面对这这么多的路，他想找一条最安全的路。但是8600 的数学不好，想请你帮忙 ^_^
Input 输入包括多个测试实例，每个实例包括：
Output 如果86无法达到他的目的地，输出"What a pity!",
Sample Input 3 1 0.5 0.5 0.5 1 0.4 0.5 0.4 1 3 1 2 2 3 1 3
Sample Output 0.500 0.400 0.500 |
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HDU TodayTime Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 41977 Accepted Submission(s): 10024 Problem Description 经过锦囊相助，海东集团终于度过了危机，从此，HDU的发展就一直顺风顺水，到了2050年，集团已经相当规模了，据说进入了钱江肉丝经济开发区500强。这时候，XHD夫妇也退居了二线，并在风景秀美的诸暨市浬浦镇陶姚村买了个房子，开始安度晚年了。
Input 输入数据有多组，每组的第一行是公交车的总数N(0<=N<=10000)；
Output 如果徐总能到达目的地，输出最短的时间；否则，输出“-1”。
Sample Input 6 xiasha westlake xiasha station 60 xiasha ShoppingCenterofHangZhou 30 station westlake 20 ShoppingCenterofHangZhou supermarket 10 xiasha supermarket 50 supermarket westlake 10 -1
Sample Output 50 Hint: The best route is: xiasha->ShoppingCenterofHangZhou->supermarket->westlake 虽然偶尔会迷路，但是因为有了你的帮助 **和**从此还是过上了幸福的生活。 ――全剧终――
Author lgx |
用map离散化一下
# if 01
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畅通工程续Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 78535 Accepted Submission(s): 30236 Problem Description 某省自从实行了很多年的畅通工程计划后，终于修建了很多路。不过路多了也不好，每次要从一个城镇到另一个城镇时，都有许多种道路方案可以选择，而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。
Input 本题目包含多组数据，请处理到文件结束。 Output 对于每组数据，请在一行里输出最短需要行走的距离。如果不存在从S到T的路线，就输出-1. Sample Input 3 3 0 1 1 0 2 3 1 2 1 0 2 3 1 0 1 1 1 2 Sample Output 2 -1 Author linle Source 2008浙大研究生复试热身赛（2）——全真模拟 |
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六度分离Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13756 Accepted Submission(s): 5537 Problem Description 1967年，美国著名的社会学家斯坦利·米尔格兰姆提出了一个名为“小世界现象(small world phenomenon)”的著名假说，大意是说，任何2个素不相识的人中间最多只隔着6个人，即只用6个人就可以将他们联系在一起，因此他的理论也被称为“六度分离”理论(six degrees of separation)。虽然米尔格兰姆的理论屡屡应验，一直也有很多社会学家对其兴趣浓厚，但是在30多年的时间里，它从来就没有得到过严谨的证明，只是一种带有传奇色彩的假说而已。
Input 本题目包含多组测试，请处理到文件结束。
Output 对于每组测试，如果数据符合“六度分离”理论就在一行里输出"Yes"，否则输出"No"。
Sample Input 8 7 0 1 1 2 2 3 3 4 4 5 5 6 6 7 8 8 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 0
Sample Output Yes Yes
Author linle |
# include
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一个人的旅行Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 56733 Accepted Submission(s): 18853 Problem 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；
Output 输出草儿能去某个喜欢的城市的最短时间。
Sample Input 6 2 3 1 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
Author Grass
Source RPG专场练习赛
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记录下想去的地方和自己家，然后找最短距离就可以
//hdu 2066
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产生冠军Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24987 Accepted Submission(s): 11174 Problem Description 有一群人，打乒乓球比赛，两两捉对撕杀，每两个人之间最多打一场比赛。
Input 输入含有一些选手群，每群选手都以一个整数n(n<1000)开头，后跟n对选手的比赛结果，比赛结果以一对选手名字（中间隔一空格）表示，前者战胜后者。如果n为0，则表示输入结束。
Output 对于每个选手群，若你判断出产生了冠军，则在一行中输出“Yes”，否则在一行中输出“No”。
Sample Input 3 Alice Bob Smith John Alice Smith 5 a c c d d e b e a d 0
Sample Output Yes No
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对map应用的水题
//hdu 2094
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ArbitrageTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11300 Accepted Submission(s): 5130 Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.
Input The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Output For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".
Sample Input 3 USDollar BritishPound FrenchFranc 3 USDollar 0.5 BritishPound BritishPound 10.0 FrenchFranc FrenchFranc 0.21 USDollar 3 USDollar BritishPound FrenchFranc 6 USDollar 0.5 BritishPound USDollar 4.9 FrenchFranc BritishPound 10.0 FrenchFranc BritishPound 1.99 USDollar FrenchFranc 0.09 BritishPound FrenchFranc 0.19 USDollar 0
Sample Output Case 1: Yes Case 2: No
Source |
用map离散一下
//hdu 1217
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Saving James BondTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4274 Accepted Submission(s): 921 Problem Description This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Input The input consists of several test cases. Each case starts with a line containing n <= 100, the number of crocodiles, and d > 0, the distance that James could jump. Then one line follows for each crocodile, containing the (x, y) location of the crocodile. Note that x and y are both integers, and no two crocodiles are staying at the same position.
Output For each test case, if James can escape, output in one line the shortest length he has to jump and the min-steps he has to jump for shortest length. If it is impossible for James to escape that way, simply ouput "can't be saved".
Sample Input 4 10 17 0 27 0 37 0 45 0 1 10 20 30
Sample Output 42.50 5 can't be saved
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//hdu 1245
#include
#include
#include
#include
using namespace std;
const double inf = 100000000;
double map[105][105];
int s[105],e[105],step[105][105],len1,len2;
struct node
{
double x,y;
} a[105];
void floyd(int n)
{
int i,j,k;
for(k = 0; k<=n; k++)
for(i = 0; i<=n; i++)
for(j = 0; j<=n; j++)
if(map[i][j]>map[i][k]+map[k][j])
{
map[i][j]=map[i][k]+map[k][j];
step[i][j]=step[i][k]+step[k][j];
}
}
int main()
{
int n,i,j,k,len;
double d,x,y;
while(~scanf("%d%lf",&n,&d))
{
len = 1;
for(i = 1; i<=n; i++)
{
scanf("%lf%lf",&x,&y);
if(fabs(x)<=7.5 && fabs(y)<=7.5)//只将所有在小岛外的点存入图中
continue;
a[len].x = x;
a[len++].y = y;
}
n = len;
if(n == 1)//只存了一个点，直接判断
{
if(d>=42.5)
printf("42.50 1\n");
else
printf("can't be saved\n");
continue;
}
for(i = 0; i<=n; i++)
for(j = 0; j<=n; j++)
{
map[i][j] = inf;
step[i][j] = 0;
}
for(i = 1; id)//步伐不能到达，将该店变为无穷大
{
map[i][j] = inf;
step[i][j] = 0;
}
}
}
len1 = len2 = 0;
for(i = 1; i Shortest PathTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6892 Accepted Submission(s): 1694 Problem Description When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.
Input The input consists of multiple test cases. For each test case, the first line contains three integers N, M and Q, where N is the number of vertices in the given graph, N≤300; M is the number of arcs, M≤100000; and Q is the number of operations, Q ≤100000. All vertices are number as 0, 1, 2, … , N - 1, respectively. Initially all vertices are unmarked. Each of the next M lines describes an arc by three integers (x, y, c): initial vertex (x), terminal vertex (y), and the weight of the arc (c). (c > 0) Then each of the next Q lines describes an operation, where operation “0 x” represents that vertex x is marked, and operation “1 x y” finds the length of shortest-path between x and y only through marked vertices. There is a blank line between two consecutive test cases.
Output Start each test case with "Case #:" on a single line, where # is the case number starting from 1.
Sample Input 5 10 10 1 2 6335 0 4 5725 3 3 6963 4 0 8146 1 2 9962 1 0 1943 2 1 2392 4 2 154 2 2 7422 1 3 9896 0 1 0 3 0 2 0 4 0 4 0 1 1 3 3 1 1 1 0 3 0 4 0 0 0
Sample Output Case 1: ERROR! At point 4 ERROR! At point 1 0 0 ERROR! At point 3 ERROR! At point 4
Source |
对于K的理解加深了是松弛了K个城市，也可以单个松弛,当前编号为k
//hdu 3631
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TravelTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5210 Accepted Submission(s): 1405 Problem Description PP loves travel. Her dream is to travel around country A which consists of N cities and M roads connecting them. PP has measured the money each road costs. But she still has one more problem: she doesn't have enough money. So she must work during her travel. She has chosen some cities that she must visit and stay to work. In City_i she can do some work to earn Ci money, but before that she has to pay Di money to get the work license. She can't work in that city if she doesn't get the license but she can go through the city without license. In each chosen city, PP can only earn money and get license once. In other cities, she will not earn or pay money so that you can consider Ci=Di=0. Please help her make a plan to visit all chosen cities and get license in all of them under all rules above.
Input The first line of input consists of one integer T which means T cases will follow.
Output If PP can visit all chosen cities and get all licenses, output "YES", otherwise output "NO".
Sample Input 2 4 5 10 1 2 1 2 3 2 1 3 2 1 4 1 3 4 2 3 1 8 5 2 5 2 3 10 1 2 1 100 1 2 10000 1 2 100000 1
Sample Output YES NO |
//先用最短路预处理完之后在爆搜就可以
//hdu 2544
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find the mincost routeTime Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8149 Accepted Submission(s): 3084 Problem Description 杭州有N个景区，景区之间有一些双向的路来连接，现在8600想找一条旅游路线，这个路线从A点出发并且最后回到A点，假设经过的路线为V1,V2,....VK,V1,那么必须满足K>2,就是说至除了出发点以外至少要经过2个其他不同的景区，而且不能重复经过同一个景区。现在8600需要你帮他找一条这样的路线，并且花费越少越好。
Input 第一行是2个整数N和M（N <= 100, M <= 1000)，代表景区的个数和道路的条数。
Output 对于每个测试实例，如果能找到这样一条路线的话，输出花费的最小值。如果找不到的话，输出"It's impossible.".
Sample Input 3 3 1 2 1 2 3 1 1 3 1 3 3 1 2 1 1 2 3 2 3 1
Sample Output 3 It's impossible.
Author 8600 |
对于K的理解更加深刻，可以用来处理最小环
//hdu 1599
#include
#include
#include
using namespace std;
const int maxn = 110;
const int inf = 1000000;
int dist[maxn][maxn];
int e[maxn][maxn];
int n,m;
void initial()
{
int i;
int j;
for(i = 1 ; i <= n ; ++i)
{
for(j = 1 ; j <= n ; ++j)
{
if(i == j)
e[i][j] = 0;
else
e[i][j] = inf;
}
}
}
int floyd()
{
int i;
int j;
int k;
int mincircle = inf;
// dist = e;
for(i = 1 ; i <= n ; ++i)
{
for(j = 1 ; j <= n ; ++j)
{
dist[i][j] = e[i][j];
}
}
//根据Floyed的原理，在最外层循环做了k-1次之后，dis[i][j]则代表了i到j的路径中所有结点编号都小于k的最短路径
for(k = 1 ; k <= n ; ++k)
{
//环的最小长度为edge[i][k]+edge[k][j]+i->j的路径中所有编号小于k的最短路径长度
for(i = 1 ; i < k ; ++i)
for(j = i+1 ; j < k ; ++j)
if(dist[i][j] + e[i][k] + e[k][j] < inf)
mincircle = min(mincircle,dist[i][j] + e[j][k] + e[k][i]);
//floyd原来的部分,更新dist[i][j]
for(i = 1 ; i <= n ; ++i)
for(j = 1 ; j <= n ; ++j)
if(dist[i][j] > dist[i][k] + dist[k][j])
dist[i][j] = dist[i][k] + dist[k][j];
}
return mincircle;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
initial();
int i;
for(i = 1 ; i <= m ; ++i)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(e[a][b] > c)
e[a][b] = e[b][a] = c;
}
int ans = floyd();
if(ans != inf)
printf("%d\n",ans);
else
printf("It's impossible.\n");
}
return 0;
}