Codeforces 505B - Mr. Kitayuta's Colorful Graph (DFS)

题意

给出一个图,每条边代表颜色,下面有几个询问,问从a到b有几条颜色完全一样的路

思路

建图。我用vector G[MAXN][MAXN]存图,然后从每个点到每个点,每种颜色都DFS一遍,最后输出。

比赛的时候也是写DFS,不过写的回溯,无限WA( TДT)。后来不回溯了,直接暴力每种颜色。

代码

#include <cstdio>
#include <stack>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 100 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int cases = 0;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
vector<int> G[MAXN][MAXN];
int mp[MAXN][MAXN], x, y, vis[MAXN], v, e, cnt;
int colvis[MAXN];
 
bool DFS(int cur, int target, int col)
{
    if (cur == target) return true;
    for (int i = 1; i <= v; i++)
    {
        if (SZ(G[cur][i]) == 0 || vis[i]) continue;
        for (int j = 0; j < SZ(G[cur][i]); j++)
        {
            if (G[cur][i][j] != col) continue;
            vis[i] = 1;
            if (DFS(i, target, col)) return true;
        }
    }
    return false;
}
 
int main()
{
    //ROP;
    int i, j;
    scanf("%d%d", &v, &e);
    for (int i = 0; i < e; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        G[a][b].PB(c); G[b][a].PB(c);
    }
    for (int i = 1; i <= v; i++)
        for (int j = i + 1; j <= v; j++)
        {
            int ans = 0;
            for (int k = 1; k <= e; k++)
            {
                MS(vis, 0);
                if (DFS(i, j, k)) ans++;
                mp[i][j] = mp[j][i] = ans;
            }
        }
    int q;
    scanf("%d", &q);
    while (q--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", mp[a][b]);
    }
    return 0;
}

Powered by Jekyll and Theme by solid