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

目录

作者：@dwtfukgv
本文为作者原创，转载请注明出处：https://www.cnblogs.com/dwtfukgv/p/7126059.html

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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：

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#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; }