树&图 记录

A - Lake Counting

POJ - 2386

最最最最最基础的dfs 挂这道题为了提高AC率（糖水不等式

B - Paint it really, really dark gray

CodeForces - 717E

dfs 待会写题解

C - New Year Transportation

CodeForces - 500A

简单的模拟

D - Binary Tree Traversals

HDU - 1710

给树的先序中序输出后序

贴下代码

#include <algorithm>
#include <cstdio>
#include <stack>
using namespace std;
const int maxn = 1e3 + ;
int preorder[maxn], inorder[maxn];
stack<int> st;
int n;

void build(int l1, int r1, int l2, int r2) {
    int i, j;
    st.push(preorder[l1]);
    for (i = l2; i <= r2; i++)
        if (inorder[i] == preorder[l1]) break;
    j = l1 + (i - l2 + );
    if (j <= r1 && i +  <= r2) build(j, r1, i + , r2);
    if (l1 +  <= j -  && l2 <= i - ) build(l1 + ,j - , l2, i - );
}
int main() {
    while (~scanf("%d", &n)) {
        for (int i = ; i < n; i++) scanf("%d", &preorder[i]);
        for (int i = ; i < n; i++) scanf("%d", &inorder[i]);
        build(, n-, , n-);
        while (!st.empty()) {
            printf("%d", st.top());
            st.pop();
            if (!st.empty())putchar(' ');
        }
        puts("");
    }
    return ;
}

E - Til the Cows Come Home

POJ - 2387

裸最短路dijkstra

F - 昂贵的聘礼

POJ - 1062

最短路+限制条件 进行n次dijkstra 每次把当前的level[i]设为最小的

毕竟每次最短路肯定有个最小的

G - 畅通工程

HDU - 1232

基础并查集

H - 还是畅通工程

HDU - 1233

基础最小生成树

I - Vasya and a Tree

CodeForces - 1076E

线段树+dfs 似乎好像是主席树吗（反正我不会

待会写发题解

J - Military Problem

CodeForces - 1006E

dfs序 写题解

K - Cut 'em all!

CodeForces - 982C

贪心+dfs 写题解

L - Install Air Conditioning

HDU - 4756

最小生成树+树形dp 写题解

M - Qin Shi Huang's National Road System

HDU - 4081

最小生成树+树形dp 写题解

N - Conscription

POJ - 3723

最小生成树 写题解

O - Cow Contest

POJ - 3660

floyd传递闭包 写题解

一场contest下来要写8个题解wsl