[AtCoder ARC098] Donation| 建图 | 树型dp

Donation


input1:

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4

input2:

5 8
6 4
15 13
15 19
15 1
20 7
1 3
1 4
1 5
2 3
2 4
2 5
3 5
4 5

output2:

44

input3:

9 10
131 2
98 79
242 32
231 38
382 82
224 22
140 88
209 70
164 64
6 8
1 6
1 4
1 3
4 7
4 9
3 7
3 9
5 9
2 5

output3:

582

有n个点m条边，每一个点有两个属性，在这个点上的时候要至少有A元，然后要在这个点上支付B元，可以选择在任意时候支付
求最少要携带多少元，才能够将所有的点都支付一遍

int head[maxn];
int n,m,cnt,tot;
ll a[maxn],b[maxn],c[maxn],id[maxn];
int fa[maxn];
int lson[maxn],rson[maxn];
struct node{
	int v,nex;
}e[maxn];
void addEdge(int u,int v){
	e[cnt].v = v;
	e[cnt].nex = head[u];
	head[u] = cnt ++;
}
void init(){
	for(int i=0;i<maxn;i++) fa[i] = i,head[i] = -1;
	cnt = 0;
	tot = 0;
}
bool cmp(int x,int y){
	return a[x] < a[y];
}
int find(int x){
	if(x == fa[x]) return x;
	else return fa[x] = find(fa[x]);
}
void add(int u,int v){
	int fau = find(u);
	int fav = find(v);
	if(fau == fav) return ;
	tot ++;
	u = fau;
	v = fav;
	lson[tot] = u;
	rson[tot] = v;
	a[tot] = max(a[u],a[v]);
	b[tot] = b[u] + b[v];///cost cal
	fa[u] = tot;
	fa[v] = tot;
	fa[tot] = tot;
}
ll dp[maxn];
void dfs(int u){
	if(lson[u]){
		dfs(lson[u]);
		dfs(rson[u]);
	}else dp[u] = a[u];
	dp[u] = min(
	max(a[u]-b[lson[u]],dp[lson[u]]),
	max(a[u]-b[rson[u]],dp[rson[u]])
	);
}
int main()
{
	init();
	n = read,m = read;
	for(int i=1;i<=n;i++) a[i] = read,b[i] = read;
	for(int i=1;i<=n;i++) a[i] = max(a[i] - b[i], 0LL),id[i] = i;
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<=m;i++){
		int u = read,v = read;
		addEdge(u,v);
		addEdge(v,u);
	}
	/*for(int i=1;i<=n;i++){
		cout<<id[i]<<' ';
	}
	puts("");*/
	tot = n;
	// puts("ok");
	for(int i=1;i<=n;i++){
		int u = id[i];
		for(int j = head[u]; ~j; j = e[j].nex){
			int to = e[j].v;
			if(a[to] <= a[u]) add(u,to);
		}
	}
	dfs(tot);
	cout << dp[tot] + b[tot] <<endl;
    return 0;
}