ZOJ 3708 Density of Power Network
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题目链接:ZOJ 3708 Density of Power Network
看懂题就很简单了,给出一个数n表示有n个点,给出一个数m表示有m条边,边是无向边, 给出的边中可能存在平行边,问去除平行边后的边数与点数的比值,答案保留三位小数。
#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int MAX_N = 500 + 10;struct Line{ int u, v;};Line l[MAX_N];int main(){ int T; scanf("%d", &T); while(T--) { vector <int> v1[MAX_N]; int bus, line, u, v, _max = 0; scanf("%d%d", &bus, &line); for(int i = 0; i < line; i++) { scanf("%d", &u); l[i].u = u; _max = max(u, _max); } for(int i = 0; i < line; i++) { scanf("%d", &v); l[i].v = v; } for(int i = 0; i < line; i++) { u = l[i].u, v = l[i].v; if(find(v1[u].begin(), v1[u].end(), v) == v1[u].end() && find(v1[v].begin(), v1[v].end(), u) == v1[v].end()) v1[u].push_back(v); } int line_size = 0; for(int i = 1; i <= _max; i++) line_size += v1[i].size(); double res = (line_size * 1.0) / (bus * 1.0); printf("%.3lf\n", res); } return 0;}
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