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∑k−1i=1c[i]to∑ki=1c[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;
}