zoukankan      html  css  js  c++  java
  • (模板)解决带负权最短路径 Bellman-ford 与 SPFA(前者的队列优化)

    // 模板记录
    // Bellman-ford O(V*E) s到每个点的最短路
    1
    #include<cstdio> 2 #define INF 0x3f3f3f3f 3 const int MAXN = 100+5; 4 const int MAXM = 10000+5; 5 6 struct node { 7 int u, v, w; 8 } edge[MAXM]; 9 10 int n, m, x;// 点的数量,边的数量,源点 11 int dis[MAXN]; 12 13 bool Bellman_ford() { 14 for(int i = 0; i < n-1; ++i) { 15 for(int j = 0; j != m; ++j) { 16 if(dis[edge[j].v] > dis[edge[j].u] + edge[j].w) { 17 dis[edge[j].v] = dis[edge[j].u] + edge[j].w; 18 } 19 } 20 } 21 bool flag = true; 22 for(int i = 0; i != m; ++i) { 23 if(dis[edge[i].v] > dis[edge[i].u] + edge[i].w) { 24 flag = false; 25 break; 26 } 27 } 28 return flag; 29 } 30 31 int main() { 32 cin >> n >> m >> x; 33 for(int i = 1; i <= n; ++i) dis[i] = INF; 34 dis[x] = 0; 35 for(int i = 0; i != m; ++i) { 36 cin >> edge[i].u >> edge[i].v >> edge[i].w; 37 if(edge[i].u == x) { 38 dis[edge[i].v] = edge[i].w; 39 } 40 } 41 if(Bellmam_ford()) cout << "have minimun road "; 42 else cout << "not have minimun road "; 43 return 0; 44 }

    // poj 1860 Bellman-ford 

     1 /*
     2  * @Promlem: 
     3  * @Time Limit: ms
     4  * @Memory Limit: k
     5  * @Author: pupil-XJ
     6  * @Date: 2019-10-25 16:51:13
     7  * @LastEditTime: 2019-10-25 20:25:55
     8  */
     9 #include<cstdio>
    10 #include<cstring>
    11 #include<cmath>
    12 #include<iostream>
    13 #include<string>
    14 #include<algorithm>
    15 #include<iomanip>
    16 #include<vector>
    17 #include<queue>
    18 #include<stack>
    19 #include<set>
    20 #include<map>
    21 #define read(n) n = read_n()
    22 #define rep(i, n) for(int i=0;i!=n;++i)
    23 #define per(i, n) for(int i=n-1;i>=0;--i)
    24 #define Rep(i, sta, n) for(int i=sta;i!=n;++i)
    25 #define rep1(i, n) for(int i=1;i<=n;++i)
    26 #define per1(i, n) for(int i=n;i>=1;--i)
    27 #define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
    28 #define L k<<1
    29 #define R k<<1|1
    30 #define mid (tree[k].l+tree[k].r)>>1
    31 #define eps 1e-10
    32 using namespace std;
    33 const int INF = 0x3f3f3f3f;
    34 typedef long long ll;
    35 
    36 inline int read_n() {
    37     char c=getchar();int x=0,f=1;
    38     while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    39     while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    40     return x*f;
    41 }
    42 // -----------------------------------------------------
    43 const int MAXN = 100+5;
    44 const int MAXM = 200+5;
    45 
    46 int n, m, m2, x;
    47 double sta;
    48 
    49 struct node {
    50     int u, v;
    51     double r, c;
    52 } edge[MAXM];
    53 
    54 double dis[MAXN];
    55 
    56 bool Bellman_ford() {
    57     bool flag;
    58     rep(i, n-1) {
    59         flag = false;
    60         rep(j, m) {
    61             if(dis[edge[j].v] < (dis[edge[j].u]-edge[j].c)*edge[j].r) {
    62                 dis[edge[j].v] = (dis[edge[j].u]-edge[j].c)*edge[j].r;
    63                 flag = true;
    64             }
    65         }
    66         if(!flag) break;
    67     }
    68     flag = false;
    69     rep(i, m) {
    70         if(dis[edge[i].v] < (dis[edge[i].u]-edge[i].c)*edge[i].r) {
    71             return true;
    72         }
    73     }
    74     return false;
    75 }
    76 
    77 int main() {
    78     ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    79     cin >> n >> m2 >> x >> sta;
    80     m = 0;
    81     rep(i, m2) {
    82         cin >> edge[m].u >> edge[m].v;
    83         cin >> edge[m].r >> edge[m].c;
    84         ++m;
    85         edge[m].u = edge[m-1].v;
    86         edge[m].v = edge[m-1].u;
    87         cin >> edge[m].r >> edge[m].c;
    88         ++m;
    89     }
    90     rep1(i, n) dis[i] = 0;
    91     dis[x] = sta;
    92     if(Bellman_ford()) cout << "YES
    ";
    93     else cout << "NO
    ";
    94     return 0;
    95 }
    View Code
     // SPFA O(k*E)k为平均入队列次数,稀疏图通常小于2。s到每个点的最短路
    1
    #include<iostream> 2 #include<queue> 3 using namespace std; 4 #define INF 0x3f3f3f3f 5 const int MAXN = 100+5; 6 const int MAXM = 1000+5; 7 8 int num; 9 int head[MAXN]; 10 struct node {// 链式向前星 11 int v, next; 12 int w; 13 } edge[MAXM]; 14 15 inline void add(int x, int y, int w) { 16 edge[num].v = y; 17 edge[num].next = head[x]; 18 edge[num].w = w; 19 head[x] = num++; 20 } 21 22 int n, m; 23 int dis[MAXN], vis[MAXN], inq[MAXN]; 24 25 bool SPFA(int s) { 26 queue<int> q; 27 for(int i = 1; i <= n; ++i) { 28 dis[i] = INF; 29 vis[i] = inq[i] = 0; 30 } 31 dis[i] = 0; 32 vis[s] = inq[s] = 1; 33 q.push(s); 34 while(!q.empty()) { 35 int u = q.front(); 36 u.pop(); 37 vis[u] = 0; 38 for(int i = head[x]; i != -1; i = edge[i].next) { 39 int v = edge[i].v; 40 if(dis[v] > dis[u] + edge[i].w) { 41 dis[v] = dis[u] + edge[i].w; 42 if(!vis[v]) { 43 q.push(v); 44 vis[v] = 1; 45 ++inq[v]; 46 if(inq[v] > n) return false; 47 } 48 } 49 } 50 } 51 return true; 52 } 53 54 int main() { 55 cin >> n >> m; 56 int s, e, v; 57 num = 0; 58 for(int i = 1; i <= n; ++i) head[i] = -1; 59 for(int i = 0; i != m; ++i) { 60 cin >> s >> e >> v; 61 add(s, e, v); 62 } 63 SPFA(1); 64 return 0; 65 }

     // poj 3259

      1 /*
      2  * @Promlem: 
      3  * @Time Limit: ms
      4  * @Memory Limit: k
      5  * @Author: pupil-XJ
      6  * @Date: 2019-10-26 00:52:36
      7  * @LastEditTime: 2019-10-26 01:40:33
      8  */
      9 #include<cstdio>
     10 #include<cstring>
     11 #include<cmath>
     12 #include<iostream>
     13 #include<string>
     14 #include<algorithm>
     15 #include<iomanip>
     16 #include<vector>
     17 #include<queue>
     18 #include<stack>
     19 #include<set>
     20 #include<map>
     21 #define read(n) n = read_n()
     22 #define rep(i, n) for(int i=0;i!=n;++i)
     23 #define per(i, n) for(int i=n-1;i>=0;--i)
     24 #define Rep(i, sta, n) for(int i=sta;i!=n;++i)
     25 #define rep1(i, n) for(int i=1;i<=n;++i)
     26 #define per1(i, n) for(int i=n;i>=1;--i)
     27 #define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
     28 #define L k<<1
     29 #define R k<<1|1
     30 #define mid (tree[k].l+tree[k].r)>>1
     31 #define eps 1e-10
     32 using namespace std;
     33 const int INF = 0x3f3f3f3f;
     34 typedef long long ll;
     35 
     36 inline int read_n() {
     37     char c=getchar();int x=0,f=1;
     38     while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
     39     while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
     40     return x*f;
     41 }
     42 // -----------------------------------------------------
     43 const int MAXN = 500+5;
     44 const int MAXM = 5200+5;
     45 
     46 int n, m1, m2;
     47 
     48 int num;
     49 struct node {
     50     int v, next;
     51     int w;
     52 } edge[MAXM];
     53 
     54 int head[MAXN];
     55 inline void add(int x, int y, int w) {
     56     edge[num].v = y;
     57     edge[num].next = head[x];
     58     edge[num].w = w;
     59     head[x] = num++;
     60 }
     61 
     62 int dis[MAXN], vis[MAXN], inq[MAXN];
     63 
     64 bool SPFA(int s) {
     65     queue<int> q;
     66     rep1(i, n) {
     67         dis[i] = INF;
     68         vis[i] = inq[i] = 0;
     69     }
     70     dis[s] = 0;
     71     vis[s] = inq[s] = 1;
     72     q.push(s);
     73     while(!q.empty()) {
     74         int u = q.front();
     75         q.pop();
     76         vis[u] = 0;
     77         for(int i = head[u]; i != -1; i = edge[i].next) {
     78             int v = edge[i].v;
     79             if(dis[v] > dis[u] + edge[i].w) {
     80                 dis[v] = dis[u] + edge[i].w;
     81                 if(!vis[v]) {
     82                     q.push(v);
     83                     vis[v] = 1;
     84                     ++inq[v];
     85                     if(inq[v] > n) return true;
     86                 }
     87             }
     88         }
     89     }
     90     return false;
     91 }
     92 
     93 int main() {
     94     //ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
     95     int T = read_n();
     96     int s, e, v;
     97     while(T--) {
     98         read(n); read(m1); read(m2);
     99         rep1(i, n) head[i] = -1;
    100         num = 0;
    101         rep(i, m1) {
    102             read(s); read(e); read(v);
    103             add(s, e, v);
    104             add(e, s, v);
    105         }
    106         rep(i, m2) {
    107             read(s); read(e); read(v);
    108             add(s, e, -v);
    109         }
    110         if(SPFA(1)) cout << "YES
    ";
    111         else cout << "NO
    ";
    112     }
    113     return 0;
    114 }
    View Code
  • 相关阅读:
    SVN补充
    java面试资料总结
    JAVABEAN EJB POJO区别
    Hibernate使用原生sql语句
    Discrete Logging(poj 2417)
    卡牌游戏(bzoj 3191)
    Activation(hdu 4089)
    Aeroplane chess(hdu 4405)
    LOOPS(hdu 3853)
    巧克力(zoj 1363)
  • 原文地址:https://www.cnblogs.com/pupil-xj/p/11741662.html
Copyright © 2011-2022 走看看