Codeforces 400D Dima and Bacteria 【并查集 + 最短路】

题目链接：Codeforces 400D Dima and Bacteria

题意：有$n$个点$m$条边的无向带权图。我们假设有$k$个集合，集合$i$元素个数记为$c[i]$，集合里面任意两点间距离为$0$。问我们的假设是否成立，成立的话输出$k * k$矩阵代表集合间的最短路。
注意有一个重要信息——
集合$k$里面的点编号从$\sum_{i=1}^{k-1}c[i] \;\;to \;\;\sum_{i=1}^kc[i]$，这样这道题就是大水题了。

思路：用个并查集维护就好了，最后跑一次$floyd$就$OK$了。

$AC$代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e5 + 10;
const int pN = 1e6;// <= 10^7
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int c[510];
int u[MAXN], v[MAXN], w[MAXN];
int father[MAXN], id[MAXN];
int Map[510][510];
int Find(int p) {
    int t, child = p;
    while(p != father[p]) p = father[p];
    while(child != p) {t = father[p]; father[child] = p; child = t;}
    return p;
}
void Merge(int x, int y)
{
    int fx = Find(x);
    int fy = Find(y);
    if(fx != fy)
        father[fx] = fy;
}
int main()
{
    int n, m, k; cin >> n >> m >> k;
    c[0] = 0;
    for(int i = 1; i <= k; i++) {
        cin >> c[i];
        c[i] += c[i-1];
    }
    for(int i = 0; i < m; i++) {
        cin >> u[i] >> v[i] >> w[i];
    }
    for(int i = 1; i <= n; i++) {
        father[i] = i; 
    }
    for(int i = 0; i < m; i++) {
        if(w[i] == 0) {
            Merge(u[i], v[i]);
        }
    }
    bool flag = true; c[0] = 0;
    for(int i = 1; i <= k; i++) {
        int root = Find(c[i-1]+1);
        for(int j = c[i-1]+1; j <= c[i]; j++) {
            if(Find(j) != root) {
                flag = false;
                break;
            }
            id[j] = i;
        }
        if(!flag) break;
    }
    if(!flag) {
        cout << "No" << endl;
        return 0;
    }
    cout << "Yes" << endl;
    for(int i = 1; i <= k; i++) {
        for(int j = 1; j <= k; j++) {
            if(i == j) Map[i][j] = 0;
            else Map[i][j] = INF;
        }
    }
    for(int i = 0; i < m; i++) {
        int s = id[u[i]]; int t = id[v[i]];
        Map[s][t] = Map[t][s] = min(Map[s][t], w[i]);
    }
    for(int p = 1; p <= k; p++) {
        for(int i = 1; i <= k; i++) {
            for(int j = 1; j <= k; j++) {
                Map[i][j] = min(Map[i][j], Map[i][p] + Map[p][j]);
            }
        }
    }
    for(int i = 1; i <= k; i++) {
        for(int j = 1; j <= k; j++) {
            if(j > 1) cout << " ";
            if(Map[i][j] == INF) cout << -1;
            else cout << Map[i][j];
        }
        cout << endl;
    }
    return 0;
}