CodeForces 400D Dima and Bacteria (并查集+Floyd)

题意：给出 n 个细菌，m 种仪器，细菌有 k 种，每种细菌数量 c[i]，给出从第 ui 细菌到第 vi 个细菌转化需要的花费。判断同种细菌之间的转化是不是花费都可以是0，如果可以再输出不同种细菌之间转化的最小花费。

析：首先要判断是同种细菌是不是转化花费为0，如果数据小的话，可以用Floyd，但是数据太大，我们可以考虑用并查集，如果在两种细菌之间转化花费为0，那么我们就用并查集将它们连接起来，然后再检查同种细菌之间转化是不是可以为0，直接用并查集判断是不是在同一集合就好，最后再求不同细菌之间转化的最小花费，因为要想不同细菌之间转化最少，那么必然是第 i 种细菌中的第 u 个细菌和第 j 种细菌中第 v 个细菌直接相连，由于数据较小，可以用Floyd，来求解。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int p[maxn];

int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
int c[maxn];

int dp[510][510];

int main(){
  int K;
  scanf("%d %d %d", &n, &m, &K);
  for(int i = 1; i <= n; ++i)  p[i] = i;
  memset(dp, INF, sizeof dp);
  for(int i = 1; i <= K; ++i){
    scanf("%d", c+i);
    c[i] += c[i-1];
    dp[i][i] = 0;
  }
  for(int i = 0; i < m; ++i){
    int u, v, val;
    scanf("%d %d %d", &u, &v, &val);
    if(val == 0){
      int x = Find(u);
      int y = Find(v);
      if(x != y)  p[y] = x;
    }
    int pos1 = lower_bound(c+1, c+1+K, u) - c;
    int pos2 = lower_bound(c+1, c+1+K, v) - c;
    dp[pos1][pos2] = dp[pos2][pos1] = min(dp[pos1][pos2], val);
  }
  bool ok = true;
  int cnt = 1;
  int x = Find(1);
  for(int i = 2; i <= n && ok; ++i){
    if(i <= c[cnt]){
      int y = Find(i);
      if(x != y)  ok = false;
    }
    else{
      x = Find(c[++cnt]);
      int y = Find(i);
      if(x != y)  ok = false;
    }
  }
  if(!ok){ puts("No");  return 0; }

  for(int k = 1; k <= K; ++k)
    for(int i = 1; i <= K; ++i)
      for(int j = 1; j <= K; ++j)
        if(dp[i][k] != INF && dp[k][j] != INF)
          dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);

  puts("Yes");
  for(int i = 1; i <= K; ++i)
    for(int j = 1; j <= K; ++j)
      if(j == K)  printf("%d\n", dp[i][j] == INF ? -1 : dp[i][j]);
      else  printf("%d ", dp[i][j] == INF ? -1 : dp[i][j]);
  return 0;
}