CodeForces - 767C Garland dfs(假剪枝)
2017-11-09 21:51
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题意:
给定一棵树,每个结点有权值,切断两条边,使剩下的3部分 每个结点的权值和相同
思路:
首先记录所有多有结点的和sum,看是否能被3除尽,
然后进行 DFS 每个结点,回溯的过程如果遇到哪个结点的值变成 sum/3 ,记录这个结点,更新这个结点的值为0 并返回;
最后还要在判断一遍(感觉不需要了)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 7, maxd = (1<<18)-1, mod = 1e9 + 7;
const int INF = 0x7f7f7f7f;
int n, sum, tt, ans1 = 0, ans2 = 0;
int d[maxn];
vector<int> a[maxn];
void init() {
memset(d, 0, sizeof d);
memset(a, 0, sizeof a);
for(int i = 1; i <= n; ++i) {
int u, t;
scanf("%d %d", &u, &t);
a[u].push_back(i);
d[i] = t; sum += t;
}
}
int dfs(int id) {
if(a[id].size() == 0) {
if(d[id] == tt) {
if(!ans1) {
ans1 = id;
d[id] = 0;
return 0;
}
else if(!ans2) {
ans2 = id;
d[id] = 0;
return 0;
}
else return d[id];
}
else return d[id];
}
int ans = 0;
for(int i = 0; i < a[id].size(); ++i) {
ans += dfs(a[id][i]);
}
d[id] += ans;
if(d[id] == tt) {
if(!ans1) {
ans1 = id;
d[id] = 0;
return d[id];
}
else if(!ans2) {
ans2 = id;
d[id] = 0;
return d[id];
}
else return d[id];
}
else return d[id];
}
void solve() {
if(sum % 3 != 0) {
puts("-1"); return;
}
tt = sum / 3;
int u = a[0][0];
d[u] = dfs(u);
if(ans1 && ans2 && ans1 != u && ans2 != u) {
4000
if(ans1 > ans2) swap(ans1, ans2);
cout << ans1 << " " << ans2 << endl;
return;
}
for(int i = 1; i <= n; ++i) {
if(i == u) continue;
if(d[i] == tt) {
if(!ans1) {
ans1 = i;
}
else if(!ans2) {
ans2 = i;
break;
}
}
}
if(ans1 && ans2 && ans1 != u && ans2 != u) {
if(ans1 > ans2) swap(ans1, ans2);
cout << ans1 << " " << ans2 << endl;
return;
}
else puts("-1");
}
int main() {
scanf("%d", &n);
init();
solve();
return 0;
}
给定一棵树,每个结点有权值,切断两条边,使剩下的3部分 每个结点的权值和相同
思路:
首先记录所有多有结点的和sum,看是否能被3除尽,
然后进行 DFS 每个结点,回溯的过程如果遇到哪个结点的值变成 sum/3 ,记录这个结点,更新这个结点的值为0 并返回;
最后还要在判断一遍(感觉不需要了)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 7, maxd = (1<<18)-1, mod = 1e9 + 7;
const int INF = 0x7f7f7f7f;
int n, sum, tt, ans1 = 0, ans2 = 0;
int d[maxn];
vector<int> a[maxn];
void init() {
memset(d, 0, sizeof d);
memset(a, 0, sizeof a);
for(int i = 1; i <= n; ++i) {
int u, t;
scanf("%d %d", &u, &t);
a[u].push_back(i);
d[i] = t; sum += t;
}
}
int dfs(int id) {
if(a[id].size() == 0) {
if(d[id] == tt) {
if(!ans1) {
ans1 = id;
d[id] = 0;
return 0;
}
else if(!ans2) {
ans2 = id;
d[id] = 0;
return 0;
}
else return d[id];
}
else return d[id];
}
int ans = 0;
for(int i = 0; i < a[id].size(); ++i) {
ans += dfs(a[id][i]);
}
d[id] += ans;
if(d[id] == tt) {
if(!ans1) {
ans1 = id;
d[id] = 0;
return d[id];
}
else if(!ans2) {
ans2 = id;
d[id] = 0;
return d[id];
}
else return d[id];
}
else return d[id];
}
void solve() {
if(sum % 3 != 0) {
puts("-1"); return;
}
tt = sum / 3;
int u = a[0][0];
d[u] = dfs(u);
if(ans1 && ans2 && ans1 != u && ans2 != u) {
4000
if(ans1 > ans2) swap(ans1, ans2);
cout << ans1 << " " << ans2 << endl;
return;
}
for(int i = 1; i <= n; ++i) {
if(i == u) continue;
if(d[i] == tt) {
if(!ans1) {
ans1 = i;
}
else if(!ans2) {
ans2 = i;
break;
}
}
}
if(ans1 && ans2 && ans1 != u && ans2 != u) {
if(ans1 > ans2) swap(ans1, ans2);
cout << ans1 << " " << ans2 << endl;
return;
}
else puts("-1");
}
int main() {
scanf("%d", &n);
init();
solve();
return 0;
}
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