hdu 2433 spfa

Travel

TimeLimit: 10000/2000 MS (Java/Others)    Memory Limit:32768/32768 K (Java/Others)
Total Submission(s): 2580    Accepted Submission(s): 845

Problem Description

      Oneday, Tom traveled to a country named BGM. BGM is a small country, but there areN (N <= 100) towns in it. Each town products one kind of food, the food willbe transported to all the towns. In addition, the trucks will always take theshortest way. There are M (M <= 3000) two-way roads connecting the towns,and the length of the road is 1.
      Let SUM be the total distance of theshortest paths between all pairs of the towns. Please write a program tocalculate the new SUM after one of the M roads is destroyed.

 

 

Input

      Theinput contains several test cases.
      The first line contains two positiveintegers N, M. The following M lines each contains two integers u, v, meaningthere is a two-way road between town u and v. The roads are numbered from 1 toM according to the order of the input.
      The input will be terminated by EOF.

 

 

Output

      OutputM lines, the i-th line is the new SUM after the i-th road is destroyed. If thetowns are not connected after the i-th road is destroyed, please output “INF”in the i-th line.

 

 

Sample Input

54

51

13

32

54

22

12

12

 

 

Sample Output

INF

INF

INF

INF

2

2

 

 

Source

2008 AsiaChengdu Regional Contest Online

 

 

题目是要求每次删除地i条边后以1,2,3,...n为起始点到其他点的最短路和的总和

分析:从题意可以想到用spfa,毕竟spfa是求单源多点的最短路的好算法, 所以可以想到每次删除第i条边时，我们只需要暂时把这条边长度赋值无穷大,然后对1,2,3,....n各点进行spfa然后求和,这是最初的思想,但是 看复杂度m*n*m=9*10^8肯定会超时,第一个m代表删除m条边,第二个n表示每次删除一条边后就对1,2,3....n个点进行spfa,第三个 m表示spfa的复杂度(由于边长度为1所以每个点其实只要入队一次,所以复杂度是m),在m*n*m中我们能优化的只能是第一个m了,根据上面n表示的 意思我们知道每起始个点都要因为m条边而进行spfa
m次,但是其实以起始点i开始的最短路边只有n-1,条,也就是说有m-n+1条边与以起始点i开始的最短路无关(即影响不到),那么如何去判断什么时候进行spfa呢??在这里引进一个数组flag[i][u][v]表示以点i作为起始点的最短路是否含有边u-v,所以我们要预先进行spfa来标记flag

这样就能根据flag[i][u[v]]来决定什么时候进行spfa,所以复杂度就变成了n*n*m,第一个n表示每个点最多要进行n-1(o(n))次

AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX = 100+10;int dist[MAX], Edgenum[MAX][MAX];bool mark[MAX],flag[MAX][MAX][MAX];int size,head[MAX],n,m,pos[3002],sum[MAX];struct node{    int v,w;    int next;    node(){}    node(int V,int W,int Next):v(V),w(W),next(Next){}}edge[6500];void Init(int num){    memset(head, -1, sizeof(int)*(num + 2));    memset(Edgenum, 0, sizeof Edgenum);    memset(flag, false, sizeof(flag));    size = 0;}void put_edge(int u,int v,int w){    edge[size] = node(v,w,head[u]);    head[u] = size++;}int spfa(int s,bool p){    int ans = 0;    queue<int>enqueue;    for(int i = 1; i <= n; ++i)    {            dist[i] = INF;            mark[i] = false;    }    dist[s] = 0;    mark[s] = true;    enqueue.push(s);    while(!enqueue.empty())    {        int u = enqueue.front();        enqueue.pop();        mark[u] = false;        for(int j = head[u]; j !=-1; j = edge[j].next)        {            int v = edge[j].v;            if(dist[v] > dist[u] + edge[j].w)            {                dist[v] = dist[u] + edge[j].w;                if(!mark[v])                {                    if(p)                        flag[s][u][v] = flag[s][v][u] = true;                    enqueue.push(v);                    mark[v] = true;                }            }        }    }    for(int i = 1; i <= n; ++i)    {        if(dist[i] == INF)            return INF;        else            ans+=dist[i];    }    return ans;}int main(){    int ans,u,v,temp;    while(cin >> n >> m)    {        Init(n);        for(int i = 0; i < m; ++i)        {            cin >> u >> v;            pos[i] = size;            put_edge(u,v,1);            put_edge(v,u,1);            ++Edgenum[u][v];            ++Edgenum[v][u];        }        ans = temp = 0;        for(int i = 1; i <= n; ++i)        {            sum[i] = spfa(i,1);            if(sum[i] == INF)            {                ans = INF;                break;            }            else                ans+=sum[i];        }        for(int i = 0; i < m; ++i)        {            u = edge[pos[i]+1].v;            v = edge[pos[i]].v;            int s = ans;            if(ans == INF)                cout << "INF" << endl;            else if(Edgenum[u][v] - 1 > 0)                cout << ans << endl;            else            {                edge[pos[i]+1].w = INF;                edge[pos[i]].w= INF;                for(int j = 1; j <= n; ++j)                {                    if(flag[j][u][v])                    {                        temp = spfa(j,0);                        if(temp == INF)                        {                            cout << "INF" << endl;                            break;                        }                        else                            s+=temp-sum[j];                    }                }                if(temp != INF)                    cout << s << endl;                edge[pos[i]+1].w = 1;                edge[pos[i]].w = 1;            }        }    }    return 0;}