UESTC Training for Graph Theory——I、War

War

Time Limit: 4000 ms Memory Limit: 65536 kB Solved: 57 Tried: 690

Description

Your country is now involved in a war! In the front, there are N positions between which you can transfer goods. Unfortunately the enemies will attack you and destroy some positions. When transferring, you cannot pass through a position which has been destroyed. As the commander of the army, you want to know the minimum distance between some position u and v, after some positions have been destroyed.

Input

The first line of inputs gives t (t <= 15) indicating the number of test cases.
Then t blocks follow. The first line of each block gives n and m (2 <= n <= 200, 0 <= m <= 100000) indicating the number of positions and number of roads. Each of the next m lines contains 3 integers u, v and e (1 <= u, v <= n, 1 <= e <= 1000), telling that there is a road between position u and position v and the distance of the road is e. An integer q (q<=10000) follows in the next line representing the number of queries to process. Each of the next q lines has one of the following formats:
A. 0 u v: This means you should reply with the shortest path between position u and position v.
B. 1 u: This means position u is destroyed.

You should notice that all the positions are not destroyed initially and there could be more than one road between two positions.

Output

For all the queries in the format A:0 u v, if any of the queried position is destroyed or there is no path from u to v, print -1; otherwise print the shortest path between the two positions. Print a blank line after each test case.

Sample Input

1
4 4
1 2 1
1 3 3
2 4 2
3 4 4
5
0 1 4
1 2
0 1 4
1 3
0 1 4

Sample Output

3
7
-1

Source

经典问题

 

/*算法思想：  给一个带权图，然后有一些操作：  1、删掉每个点  2、询问两点之间的最短路  变相的floyd，灵活运用了floyd的思想，先把所有的操作读入，倒着处理，先把要删掉的点全部删掉  然后用还存在在图中的点跑一遍floyd，倒着处理每个操作，要是这个操作是询问两点之间的最短路，  直接给出答案，要是操作是删掉某个点，就把这个点加进来，先更新其他点到这个点的最短路，然后  再用这个点去更新所有点对之间的最短路。*/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define INF 123456789using namespace std;struct data{    int bj;    int v1,v2,ans;} query[20000];int n,m;int g[300][300];bool out[300];int main(){    int t,a,b,c;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                if(i!=j) g[i][j]=INF;                else g[i][j]=0;        for(int  i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&c);            if(g[a][b]>c) g[a][b]=g[b][a]=c;        }        int Q;        scanf("%d",&Q);        memset(out,0,sizeof(out));        for(int i=1;i<=Q;i++)        {            scanf("%d",&query[i].bj);            if(query[i].bj)            {                scanf("%d",&query[i].v1);                out[query[i].v1]=true;            }            else scanf("%d%d",&query[i].v1,&query[i].v2);        }        for(int k=1;k<=n;k++)  //把所有的点删掉之后用还在图中的点跑一遍floyd        if(!out[k])            for(int i=1;i<=n;i++)            if(!out[i])                for(int j=1;j<=n;j++)                if(!out[j])                    g[i][j]=min(g[i][j],g[i][k]+g[k][j]);        for(int i=Q;i>=1;i--)        {            if(query[i].bj)            {                int k=query[i].v1;                out[k]=false;                for(int i=1;i<=n;i++)  //先更新其他点到这个点的最短路                if(!out[i])                    for(int j=1;j<=n;j++)                    if(!out[j])                        g[i][k]=min(g[i][k],g[i][j]+g[j][k]);                for(int i=1;i<=n;i++)                    if(!out[i]) g[k][i]=g[i][k];                for(int i=1;i<=n;i++)  //再用这个点更新所有点对之间的最短路                if(!out[i])                    for(int j=1;j<=n;j++)                    if(!out[j])                        if(g[i][j]>g[i][k]+g[k][j]) g[i][j]=g[i][k]+g[k][j];            }            else            {                if(!out[query[i].v1] && !out[query[i].v2] && g[query[i].v1][query[i].v2]<INF)                    query[i].ans=g[query[i].v1][query[i].v2];  //存储当前的最短路                else query[i].ans=-1;  //不能到达            }        }        for(int i=1;i<=Q;i++)            if(!query[i].bj)                printf("%d\n",query[i].ans);        printf("\n");    }    return 0;}