hdu Connections between cities

Connections between cities

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 393 Accepted Submission(s): 118

Problem Description

After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well. Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.

Input

Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.

Output

For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.

Sample Input

Sample Output

Not connected
6

Hint

Hint Huge input, scanf recommended.

Source

2009 Multi-University Training Contest 8 - Host by BJNU

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分析：LCA转为RMQ

#include<cstdio>
#include<cstring>
#include<cmath>
#define maxn 10010

int N, M, Q;

typedef struct S1 {
    int v, w;
    struct S1 *next;
} EDGE;
EDGE *node[maxn], edge[maxn * 2];
int tot;
int fa[maxn];

int find(int a) {
    if (a != fa[a])
        fa[a] = find(fa[a]);
    return fa[a];
}

void merge(int a, int b) {
    a = find(a);
    b = find(b);
    fa[a] = b;
}

void add(int u, int v, int w) {
    edge[++tot].v = v;
    edge[tot].w = w;
    edge[tot].next = node[u];
    node[u] = &edge[tot];
}
int first[maxn];
int rlen[maxn];
int deep[maxn * 2];
int in[maxn * 2];
int cnt;

void dfs(int root, int DEEP, int RLEN) {
    rlen[root] = RLEN;
    first[root] = ++cnt;
    deep[cnt] = DEEP;
    in[cnt] = root;
    EDGE *temp = node[root];
    while (temp) {
        if (!first[temp->v]) {
            dfs(temp->v, DEEP + 1, RLEN + temp->w);
            deep[++cnt] = DEEP;
            in[cnt] = root;
        }
        temp = temp->next;
    }
}
int d[maxn * 2][15];

int min(int a, int b) {
    return a < b ? a : b;
}

void st(int n) {
    int i, j, k;
    k = (int) (log((double) n) / log(2.0));
    for (i = 1; i <= n; ++i)
        d[i][0] = i;
    for (j = 1; j <= k; ++j) {
        for (i = 1; i <= n - (1 << j) + 1; ++i) {
            if (deep[d[i][j - 1]] < deep[d[i + (1 << (j - 1))][j - 1]])
                d[i][j] = d[i][j - 1];
            else
                d[i][j] = d[i + (1 << (j - 1))][j - 1];
        }
    }
}

int rmq(int a, int b) {
    int k;
    k = (int) (log(b - a + 1.0) / log(2.0));
    if (deep[d[a][k]] < deep[d[b - (1 << k) + 1][k]])
        return d[a][k];
    else
        return d[b - (1 << k) + 1][k];
}

int lca(int a, int b) {
    a = first[a];
    b = first[b];
    if (a < b)
        return in[rmq(a, b)];
    else
        return in[rmq(b, a)];
}
int hash[maxn];

void init() {
    int i;
    for (i = 0; i <= N; ++i) {
        fa[i] = i;
        node[i] = NULL;
    }
    tot = 0;
    memset(hash, 0, sizeof (hash));
    memset(first, 0, sizeof (first));
    cnt = 0;
}

int main() {
    int i, a, b, c;
    while (scanf("%d%d%d", &N, &M, &Q) != EOF) {
        init();
        while (M--) {
            scanf("%d%d%d", &a, &b, &c);
            add(a, b, c);
            add(b, a, c);
            merge(a, b);
        }
        for (i = 1; i <= N; ++i) {
            hash[find(i)] = 1;
            //      printf("%d fa=%d\n", i, fa[i]);
        }
        for (i = 1; i <= N; ++i) {
            if (hash[i]) {
                add(i, 0, 0);
                add(0, i, 0);
            }
        }
        dfs(0, 0, 0);
        st(cnt);
        while (Q--) {
            scanf("%d%d", &a, &b);
            if (fa[a] != fa[b])
                printf("Not connected\n");
            else
                printf("%d\n", rlen[a] + rlen[b] - 2 * rlen[lca(a, b)]);
        }
    }
    return 0;
}