洛谷P4578 [FJOI2018]所罗门王的宝藏(dfs)

题意

题目链接

sol

对于每个询问\(x, y, c\)

从在\((x, y)\)之间连一条边权为\(c\)的双向边，然后就是解\(k\)个二元方程。

随便带个数进去找找环就行了

#include<bits/stdc++.h>
#define ll long long 
#define fi first
#define se second
#define pair pair<int, int> 
#define fin(x) freopen(#x".in", "r", stdin);
using namespace std;
const int maxn = 3001, inf = 1e9 + 10;
template<typename a, typename b> inline bool chmax(a &x, b y) {return x < y ? x = y, 1 : 0;}
template<typename a, typename b> inline bool chmin(a &x, b y) {return x > y ? x = y, 1 : 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, val[maxn], vis[maxn];
vector<pair> v[maxn];
int dfs(int x) {
    vis[x] = 1;
    for(auto &to : v[x]) {
        if(vis[to.fi] && val[x] + val[to.fi] != to.se) return 0;
        else if(vis[to.fi]) continue;
        val[to.fi] = to.se - val[x];
        if(!dfs(to.fi)) return 0;
    }
    return 1;
}
void solve() {
    memset(val, 0, sizeof(val));
    memset(vis, 0, sizeof(vis));
    int n = read(), m = read(), k = read();
    for(int i = 1; i <= n + m; i++) v[i].clear();
    for(int i = 1; i <= k; i++) {
        int x = read(), y = read(), c = read();
        v[x].push_back({y + n, c});
        v[y + n].push_back({x, c});
    }
    for(int i = 1; i <= n + m; i++) 
        if(!vis[i] && !dfs(i)) 
            {puts("no"); return ;}
    puts("yes");
}
signed main() {
//  fin(a);
    for(int t = read(); t--; solve());
    return 0;
}