oyjpArt ACM/ICPC算法程序设计空间

Sightseeing trip
Time Limit:1000MS  Memory Limit:65536K
Total Submit:317 Accepted:133 Special Judged

Description
There is a travel agency in Adelton town on Zanzibar island. It has decided to offer its clients, besides many other attractions, sightseeing the town. To earn as much as possible from this attraction, the agency has accepted a shrewd decision: it is necessary to find the shortest route which begins and ends at the same place. Your task is to write a program which finds such a route.

In the town there are N crossing points numbered from 1 to N and M two-way roads numbered from 1 to M. Two crossing points can be connected by multiple roads, but no road connects a crossing point with itself. Each sightseeing route is a sequence of road numbers y_1, ..., y_k, k>2. The road y_i (1<=i<=k-1) connects crossing points x_i and x_{i+1}, the road y_k connects crossing points x_k and x_1. All the numbers x_1,...,x_k should be different.The length of the sightseeing route is the sum of the lengths of all roads on the sightseeing route, i.e. L(y_1)+L(y_2)+...+L(y_k) where L(y_i) is the length of the road y_i (1<=i<=k). Your program has to find such a sightseeing route, the length of which is minimal, or to specify that it is not possible,because there is no sightseeing route in the town.

Input
The first line of input contains two positive integers: the number of crossing points N<=100 and the number of roads M<=10000. Each of the next M lines describes one road. It contains 3 positive integers: the number of its first crossing point, the number of the second one, and the length of the road (a positive integer less than 500).

Output
There is only one line in output. It contains either a string 'No solution.' in case there isn't any sightseeing route, or it contains the numbers of all crossing points on the shortest sightseeing route in the order how to pass them (i.e. the numbers x_1 to x_k from our definition of a sightseeing route), separated by single spaces. If there are multiple sightseeing routes of the minimal length, you can output any one of them.

Sample Input

5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20

Sample Output

1 3 5 2

 

点数是100个 题目意思是找一个最小权圈 以任意序输出
从图论的角度上考虑 应该是任意选一条边（枚举） 然后删除边 再以一个点为原点求Dijkstra 找出最小权圈 o(M*N^2)的复杂度 一个更加好的算法是限定枚举的点为圈内序号最大的点 这样就避免了对一个圈的多次枚举(参考程序3)
如果直接搜就是任选一个点开始走回到原点则记录长度 搜的时候必须要先对每个点的边按照边权进行排序 以备后面大量剪枝

程序1

 1//Solution
 2//by oyjpArt
 3//Algorithm:Search
 4#include <vector>
 5#include <iostream>
 6#include <algorithm>
 7using namespace std;
 8
 9const int N = 101;
10struct Node {int x, w; void set(int xx, int ww) {x =xx; w = ww; }};
11vector<Node> adj[N];
12int nv, ne, ans[N], na, S, rec[N];
13bool chk[N];
14int best;
15
16bool operator<(const Node& a, const Node& b) {
17 return a.w < b.w;
18}
19
20void search(int x, int sum, int depth, int father) {
21 int i;
22 if(x == S && chk[x]) {
23  if(sum < best) {
24   best = sum;  na = depth;
25   for(i = 0; i < depth; i++) ans[i] = rec[i];
26  }
27  return;
28 }
29 rec[depth] = x;
30 for(i = 0; i < adj[x].size(); ++i) if(adj[x][i].x != father) if(!chk[adj[x][i].x] || adj[x][i].x == S) {
31  chk[adj[x][i].x] = 1;
32  if(sum + adj[x][i].w < best) search(adj[x][i].x, sum + adj[x][i].w, depth+1, x);
33  chk[adj[x][i].x] = 0;
34 }
35} 
36
37int main() {
38 scanf("%d %d", &nv, &ne);
39 int i, u, v, w;
40 Node now;
41 for(i = 0; i < ne; i++) {
42  scanf("%d %d %d", &u, &v, &w);
43  --u; --v;
44  now.set(v, w);
45  adj[u].push_back(now);
46  now.x = u;
47  adj[v].push_back(now);
48 }
49 for(i = 0; i < nv; ++i) 
50  sort(adj[i].begin(), adj[i].end());
51 
52 best = 123456789;
53 for(i = 0; i < nv; ++i) {
54  memset(chk, 0, nv * sizeof(bool));
55  S = i;
56  search(i, 0, 0, -1);
57 }
58
59 if(best == 123456789) { printf("No solution.\n"); return 0; }
60 printf("%d", ans[0]+1);
61 for(i = 1; i < na; ++i) printf(" %d", ans[i]+1); putchar('\n');
62
63 return 0;
64}
65
66

程序2

 1//Solution
 2//by oyjpArt
 3//Algorithm : Enumerate + Dijkstra
 4#include <stdio.h>
 5#include <string.h>
 6
 7const int N = 101, M = 20001, MAXINT = 2000000000;
 8int ne, nv;
 9struct E {
10 int x, w; E* next;
11 void set(int xx, int ww, E* nn) {x = xx; w = ww; next = nn;}
12}e[M], * head[N];
13int best, dist[N], q[N], ans[N], pre[N], na;
14bool chk[N];
15
16void Dijk(int st, int end, int ow) {
17 memset(chk, 0, sizeof(chk));
18 memset(dist, -1, sizeof(dist));
19 int qe = 1, qs = 0, i;
20 E * p;
21 for(i = 0; i < nv; ++i) if(i != st) {
22  for(p = head[st]; p != NULL; p = p->next) {
23   if(p->x == i && p->w > 0 && (dist[i] == -1 || dist[i] > p->w ) ) 
24    dist[i] = p->w;
25  }
26  if(dist[i] == -1) dist[i] = MAXINT;
27 }
28 q[0] = st;
29 dist[st] = 0;
30 chk[st] = 1;
31 for(i = 0; i < nv; ++i) pre[i] = st;
32 pre[st] = -1;
33 while(qs < qe) {
34  int cur = q[qs++];
35  chk[cur] = 1;
36  if(ow + dist[cur] >= best) return;
37  if(cur == end) {
38   if(dist[end] + ow < best) {
39    na = 0;
40    for(i = cur; i != -1; i = pre[i]) ans[na++] = i;
41    best = dist[end] + ow;
42   }
43   return;
44  }
45  int _min = MAXINT, mini = -1;
46  for(i = 0; i < nv; i++) if(!chk[i]) {
47   if(dist[i] < _min) {
48    _min = dist[i];
49    mini = i;
50   }
51  }
52  if(mini == -1) return;
53  q[qe++] = mini;
54  for(i = 0; i < nv; ++i) if(!chk[i]) {
55   for(p = head[mini]; p != NULL; p = p->next) if(p->x == i)  break;
56   if(p == NULL) continue;
57   if(p->w > 0 && p->w + dist[mini] < dist[i]) {
58    dist[i] = p->w + dist[mini];
59    pre[i] = mini;
60   }
61  }
62 }
63}
64
65int main() {
66 scanf("%d %d", &nv, &ne);
67 memset(head, NULL, nv * sizeof(E*));
68 int i, u, v, w;
69 for(i = 0; i < ne; ++i) {
70  scanf("%d %d %d", &u, &v, &w);
71  --u; --v;
72  e[2*i].set(u, w, head[v]);
73  head[v] = &e[2*i];
74  e[2*i+1].set(v, w, head[u]);
75  head[u] = &e[2*i+1];
76 }
77 E * p, * q;
78 best = MAXINT;
79 for(i = 0; i < nv; ++i) {
80  for(p = head[i]; p != NULL; p = p->next) {
81   int w = p->w;
82   int j = p->x;
83   for(q = head[i]; q != NULL; q = q->next) if(q->x == j) q->w = -q->w;
84   for(q = head[j]; q != NULL; q = q->next) if(q->x == i) q->w = -q->w;
85   Dijk(i, j, w);
86   for(q = head[i]; q != NULL; q = q->next) if(q->x == j) q->w = -q->w;
87   for(q = head[j]; q != NULL; q = q->next) if(q->x == i) q->w = -q->w;
88  }
89 }
90 if(best == MAXINT) printf("No solution.\n");
91 else {
92  printf("%d", ans[0] + 1);
93  for(i = 1; i < na; ++i) printf(" %d", ans[i] + 1); putchar('\n');
94 }
95 return 0;
96}
97//唉 不用vector代码量增大好多。。晕倒
98

 1#include <stdio.h>
 2#include <string.h>
 3
 4const int N = 101;
 5const int MAXINT = 123456789;
 6int ne, nv;
 7int adj[N][N];
 8int pre[N][N];
 9int conn[N][N];
10int na, ans[N];
11int best;
12
13void floyd() {
14    int i, j, k, tmp, p;
15    for(k = 0; k < nv; ++k) {
16        for(i = 0; i < k; ++i) {
17            for(j = 0; j < k; ++j) if(conn[i][k] && conn[k][j] && j != i) {
18                if( (tmp = adj[i][j] + conn[k][i] + conn[j][k]) < best) {
19                    best = tmp;
20                    na = 1; ans[0] = k; p = i;
21                    while(p != -1) {
22                        ans[na++] = p;
23                        p = pre[p][j];
24                    }
25                }
26            } 
27        }
28        for(i = 0; i < nv; ++i) 
29            for(j = 0; j < nv; ++j) {
30                if(adj[i][j] > adj[i][k] + adj[k][j]) {
31                    adj[i][j] = adj[i][k] + adj[k][j];
32                    pre[i][j] = pre[i][k];
33                }
34            }
35    }
36}
37
38int main() {
39    int i, j, u, v, w;
40    memset(pre, -1, sizeof(pre));
41    scanf("%d %d", &nv, &ne);
42    for(i = 0; i < nv; ++i) {
43        for(j = i+1; j < nv; ++j) 
44            adj[i][j] = adj[j][i] = MAXINT;
45        adj[i][i] = 0;
46    }
47    for(i = 0; i < ne; ++i) {
48        scanf("%d %d %d", &u, &v, &w);
49        --u; --v;
50        if(w < adj[u][v])     
51            conn[u][v] = conn[v][u] = adj[u][v] = adj[v][u] = w;
52        pre[u][v] = v, pre[v][u] = u;
53    }
54    best = MAXINT;
55    floyd();
56    if(best == MAXINT) printf("No solution.\n");
57    else {
58        for(i = 0; i < na; ++i) {
59            printf("%d", ans[i] + 1);
60            if(i != na-1) putchar(' ');
61            else putchar('\n');
62        }
63    }
64
65    return 0;
66}
67