HDU 5441 Travel（并查集）

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1485    Accepted Submission(s): 528

Problem Description

Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are n cities and m bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is x minutes, how many pairs of city (a,b) are there that Jack can travel from city a to b without going berserk?

 

Input

The first line contains one integer T,T≤5, which represents the number of test case.

For each test case, the first line consists of three integers n,m and q where n≤20000,m≤100000,q≤5000. The Undirected Kingdom has n cities and mbidirectional roads, and there are q queries.

Each of the following m lines consists of three integers a,b and d where a,b∈{1,...,n} and d≤100000. It takes Jack d minutes to travel from city a to city band vice versa.

Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.

 

Output

You should print q lines for each test case. Each of them contains one integer as the number of pair of cities (a,b) which Jack may travel from a to b within the time limit x.

Note that (a,b) and (b,a) are counted as different pairs and a and b must be different cities.

 

Sample Input

15 5 32 3 63341 5 157243 5 57054 3 123821 3 2172660001000013000

 

Sample Output

2612

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

   题意：每2个城市间有一个权值，还有每次询问会给一个权值，城市间的权值小于询问的权值即可以走，问某个人能走多少对城市，注意(a,b),(b,a)算一对不同的城市

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;int num[21000];int v[21000];int n,m,k;int cnt;int rt[20010];struct node {    int u;    int v;    int w;} q[101000];struct node1 {    int x;    int y;    int z;} p[101000];int cmp1(const void *a,const void *b) {    struct node1 *aa = (node1*)a;    struct node1 *bb = (node1*)b;    return aa->x - bb->x;}int cmp3(const void *a,const void *b) {    struct node *aa = (node*)a;    struct node *bb = (node*)b;    return aa->w - bb->w;}void add(int u,int v,int z) {    q[cnt].u = u;    q[cnt].v = v;    q[cnt].w = z;    cnt++;}int findx(int x) {    int r = x;    while(r!=num[r]) {        r = num[r];    }    int kk = x,j;    while(kk!=r) {        j = num[kk];        num[kk] = r;        kk = j;    }    return r;}void bing(int fx,int fy,int x,int y) {    if(fx!=fy) {        num[fy] = fx;    }}int main() {    int T;    scanf("%d",&T);    while(T--) {        scanf("%d%d%d",&n,&m,&k);        int x,y,z;        cnt = 0;        for(int i=0; i<m; i++) {            scanf("%d%d%d",&x,&y,&z);            add(x,y,z);        }        int x1;        qsort(q,cnt,sizeof(q[0]),cmp3);        for(int i=0; i<k; i++) {            scanf("%d",&x1);            p[i].z = 0;            p[i].y = i;            p[i].x = x1;        }        qsort(p,k,sizeof(p[0]),cmp1);        for(int j=1; j<=n; j++) {            num[j] = j;            rt[j] = 0;            v[j] = 1;        }        y = -1;        int pj = 0;        for(int i=0; i<k; i++) {            x = p[i].x;            for(int j=pj; j<cnt; j++) {                if(q[j].w>x){                    break;                }                pj = j;                if(q[j].w>y && q[j].w<=x) {                    int py = findx(q[j].u);                    int px = findx(q[j].v);                    if(px!=py){                        bing(py,px,q[j].u,q[j].v);                    int yy = v[py];                    v[py] += v[px];                    p[i].z += ((v[py])*(v[py]-1)-(v[px])*(v[px]-1)-(yy*(yy-1)));                    v[px] = 1;                    }                }            }            rt[p[i].y] = p[i].z;            y = p[i].x;            p[i+1].z = p[i].z;        }        for(int j=0;j<k;j++){            printf("%d\n",rt[j]);        }    }    return 0;}