标签：ons 答案 pac blank head print operator style 路径
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int num[1000];
int shun(){
for (int i=15;i>=5;i--){
if (num[i]&&num[i-1]&&num[i-2]&&num[i-3]&&num[i-4]) return i;
}
return -1;
}
struct node {
int rk, num1, num2, num3;
string na;
bool operator<(const node a) const {
if (rk != a.rk) return rk > a.rk;
if (num1 != a.num1) return num1 > a.num1;
if (num2 != a.num2) return num2 > a.num2;
if (num3 != a.num3) return num3 > a.num3;
return na < a.na;
}
}b[maxn];
int n;
string na,s;
unordered_map<string,int>ma;
int main() {
//freopen("1.txt", "r", stdin);
ma["A"] = 1;
ma["2"] = 2;
ma["3"] = 3;
ma["4"] = 4;
ma["5"] = 5;
ma["6"] = 6;
ma["7"] = 7;
ma["8"] = 8;
ma["9"] = 9;
ma["10"] = 10;
ma["J"] = 11;
ma["Q"] = 12;
ma["K"] = 13;
while (cin >> n) {
for (int i = 1; i <= n; i++) {
cin >> na >> s;
b[i].na = na;
string ss;
int len = s.length();
for (int j = 0; j <= 15; j++) num[j] = 0;
for (int j = 0; j < len; j++) {
ss = "";
if (s[j] != ‘1‘) ss = s[j];
else {
ss = s[j];
ss = ss + s[j + 1];
j++;
}
num[ma[ss]]++;
}
if (num[10] && num[11] && num[12] && num[13] && num[1]) {
b[i].rk = 8;
b[i].num1 = b[i].num2 = b[i].num3 = 0;
continue;
}
int k=shun();
if (k!=-1) {
b[i].rk = 7;
b[i].num1 = k;
continue;
}
int f4 = 0, kk;
for (int j = 0; j <= 15; j++) {
if (num[j] == 4) {
kk = j;
f4 = 1;
break;
}
}
if (f4) {
int kkk;
for (int j = 0; j <= 15; j++) {
if (num[j] == 1) {
kkk = j;
break;
}
}
b[i].rk = 6;
b[i].num1 = kk;
b[i].num2 = kkk;
b[i].num3 = 0;
continue;
}
int f3 = 0;
for (int j = 0; j <= 15; j++) {
if (num[j] == 3) {
kk = j;
f3 = 1;
break;
}
}
if (f3) {
int kkk, f2 = 0;
for (int j = 0; j <= 15; j++) {
if (num[j] == 2) {
kkk = j;
f2 = 1;
break;
}
}
if (f2) {
b[i].rk = 5;
b[i].num1 = kk;
b[i].num2 = kkk;
b[i].num3 = 0;
continue;
}
int sum = 0;
for (int j = 0; j <= 15; j++) {
if (num[j] == 1) {
sum += j;
}
}
b[i].rk = 4;
b[i].num1 = kk;
b[i].num2 = sum;
b[i].num3 = 0;
continue;
}
kk = 0;
for (int j = 0; j <= 15; j++) {
if (num[j] == 2) {
kk++;
}
}
if (kk == 2) {
int mx = 0, mi = 10000, kkk;
for (int j = 0; j <= 15; j++) {
if (num[j] == 2) {
mx = max(mx, j);
mi = min(mi, j);
}
if (num[j] == 1) {
kkk = j;
}
}
b[i].rk = 3;
b[i].num1 = mx;
b[i].num2 = mi;
b[i].num3 = kkk;
continue;
}
if (kk == 1) {
int kkk, sum = 0;
for (int j = 0; j <= 15; j++) {
if (num[j] == 2) {
kkk = j;
} else if (num[j]) sum += j;
}
b[i].rk = 2;
b[i].num1 = kkk;
b[i].num2 = sum;
b[i].num3 = 0;
continue;
}
int sum = 0;
for (int j = 0; j <= 15; j++) {
if (num[j]) sum += j;
}
b[i].rk = 1;
b[i].num1 = sum;
b[i].num2 = 0;
b[i].num3 = 0;
}
sort(b + 1, b + n + 1);
for (int i = 1; i <= n; i++) {
cout << b[i].na << endl;
}
}
return 0;
}
题意：求一颗树中所有点对(a,b)的路径长度，路径长度按照模3之后的值进行分类，最后分别求每一类的和
分析：树形DP
dp[i][j]表示以 i 为根的子树中，所有子节点到 i 的路径长度模3等于 j 的路径之和
c[i][j]表示以 i 为根的子树中，所有子节点到 i 的路径长度模3等于 j 的点数
ok[i][j]表示以 i 为根的子树中，是否有子节点到 i 的路径长度模3等于 j
每次只考虑所有经过根 x 的路径，并且路径的一个端点在 x 的一颗子树上，另一个端点在 x 的另一颗子树上。（可以想到其他所有情况都可以在考虑 x 的子树结点或者是x的祖先结点时被考虑到）
假设当前枚举到 x 的子节点 y，之前遍历的子节点已经使得三个数组更新。那么我们假设要计算的路径的起点在 y ，要计算的路径的终点在之前遍历过的子节点中。
计算答案贡献：
关于x-y的连边的贡献为
c[x][a]*c[y][b]*edge
关于起点到 y 的所有路径长度的贡献为
c[x][a]*dp[y][b]
关于x到终点的所有路径长度的贡献为
c[y][b]*dp[x][a]
最终边权所属分类为(a+b+edge)%3累加到答案即可
关于更新 x
用 y 来更新 x
dp[x][(a+edge)%3]+=dp[y][a]+edge∗c[y][a]
ok[x][(a+edge)%3]=true
c[x][(a+edge)%3]+=c[y][a]
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=10020;
const int maxm=200010;
const ll mod=1e9+7;
int t,head[maxn];
struct edge{
int v,next;
ll w;
}e[maxm];
bool ok[maxn][3];
int n;
ll dp[maxn][3],c[maxn][3],ans[3];
void add(int u,int v,int w) {
t++;
e[t].v = v;
e[t].w = w;
e[t].next = head[u];
head[u] = t;
}
void dfs(int u,int fa) {
for (int i = head[u]; i; i = e[i].next) {
int v = e[i].v;
if (v == fa) continue;
dfs(v, u);
ll w = e[i].w;
for (int j = 0; j < 3; j++) {
for (int k = 0; k < 3; k++) {
if (ok[u][j] && ok[v][k]) {
ans[(j + k + w) % 3] += (dp[u][j] * c[v][k] % mod + dp[v][k] * c[u][j] % mod) % mod;
ans[(j + k + w) % 3] += w * c[u][j] % mod * c[v][k] % mod;
ans[(j + k + w) % 3] %= mod;
}
}
}
for (int j = 0; j < 3; j++) {
if (ok[v][j]) {
dp[u][(j + w) % 3] += dp[v][j] + w * c[v][j] % mod;
dp[u][(j + w) % 3] %= mod;
c[u][(j + w) % 3] += c[v][j];
ok[u][(j + w) % 3] = 1;
}
}
}
}
int main() {
while (~scanf("%d", &n)) {
t=0;
for (int i = 1; i <= n; i++) {
dp[i][0] = dp[i][1] = dp[i][2] = 0;
c[i][1] = c[i][2] = 0;
c[i][0] = 1;
ok[i][0] = 1;
ok[i][1] = ok[i][2] = 0;
head[i] = 0;
ans[0]=ans[1]=ans[2]=0;
}
for (int i = 1, u, v,w; i < n; i++) {
scanf("%d%d%d", &u, &v, &w);
u++;
v++;
add(u, v, w);
add(v, u, w);
}
dfs(1, 0);
printf("%lld %lld %lld\n", ans[0] * 2 % mod, ans[1] * 2 % mod, ans[2] * 2 % mod);
}
return 0;
}
The Preliminary Contest for ICPC Asia Shenyang 2019
标签：ons 答案 pac blank head print operator style 路径
原文地址：https://www.cnblogs.com/Accpted/p/11521281.html
