水题都做不好，看来凉了
T1 音量调节(changingsounds)
这是个小菜鸡题
只需要 (dp) 一波就行（理论上都不是 (dp)）
(dp[i][j]) 表示第 (i) 首歌曲能否达到 (j) 的音量
在 (dp[n][i]=1) 找到 (i) 最大的
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int)a.size()
#define pb push_back
#define mp make_pair
#define ll long long
#define fi first
#define se second
const int N = 60 ;
int n, b, e ;
int a[N], f[N][1010] ;
signed main() {
// freopen("changingsounds.in", "r", stdin) ;
// freopen("changingsounds.out", "w", stdout) ;
scanf("%d%d%d", &n, &b, &e) ;
rep(i, 1, n) scanf("%d", &a[i]) ;
f[0][b] = 1 ;
rep(i, 1, n)
rep(j, 0, e) {
if (j - a[i] >= 0) f[i][j] |= f[i - 1][j - a[i]] ;
if (j + a[i] <= e) f[i][j] |= f[i - 1][j + a[i]] ;
}
int ans = -1 ;
per(i, e, 0)
if (f[n][i]) {
ans = i ;
break ;
}
if (ans == -1) printf("%d
", ans) ;
else printf("%d
", ans) ;
return 0 ;
}
T2 旅行（journey）
这个题目稍难些
可以发现前50分按照他那个公式处理就行了（送的真多）
对于剩下的50分，你会发现公式是错的（doge）
会有 (X) 挡路，所有需要越过 (X)，
而每越过一个 (X)，你的距离就会增加 (2)。
由于没有连续的对角线挡路，所以只要考虑多少对 ((i,j)) 被 (X) 挡住了。
但是刚开始只想到了在同一行、列内如果挡住了会有需要绕过 (X) 的情况，
但其实如果不同的两行、列之间，所走的曼哈顿路径也被 (X) 挡住了，
那也是需要进行绕路的。
#include <bits/stdc++.h>
using namespace std ;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int) a.size()
#define pb push_back
#define mp make_pair
#define ll long long
#define fi first
#define se second
const int N = 1010;
char ch[N][N];
int x[N], y[N], n, m;
int main() {
scanf("%d%d", &n, &m);
rep(i, 0, n - 1)
rep(j, 0, m - 1)
scanf(" %c", &ch[i][j]);
rep(i, 0, n - 1) {
y[i] = -1;
rep(j, 0, m - 1) {
if (ch[i][j] == 'X') {
y[i] = j;
break;
}
}
}
rep(j, 0, m - 1) {
x[j] = -1;
rep(i, 0, n - 1) {
if (ch[i][j] == 'X') {
x[j] = i;
break;
}
}
}
ll sum = 0;
rep(i, 0, n - 1)
rep(j, i + 1, n - 1)
sum += (ll) (m - (y[i] >= 0)) * (m - (y[j] >= 0)) * (j - i) * 2;
rep(i, 0, m - 1)
rep(j, i + 1, m - 1)
sum += (ll) (n - (x[i] >= 0)) * (n - (x[j] >= 0)) * (j - i) * 2;
rep(i, 0, m - 1) {
if (x[i] == -1) continue;
int below = n - 1 - x[i], above = x[i];
sum += (ll) (below) * above * 4;
rep(j, i + 1, m - 1) {
if (x[j] == -1 || x[j] > x[j - 1]) break;
above = x[j];
sum += (ll) (below) * above * 4;
}
above = x[i];
per(j, i - 1, 0) {
if (x[j] == -1 || x[j] > x[j + 1]) break;
above = x[j];
sum += (ll) (below) * above * 4;
}
}
rep(i, 0, n - 1) {
if (y[i] == -1) continue;
int below = y[i], above = m - 1 - y[i];
sum += (ll) (below) * above * 4;
rep(j, i + 1, n - 1) {
if (y[j] == -1 || y[j] < y[j - 1]) break;
above = m - 1 - y[j];
sum += (ll) (below) * above * 4;
}
above = m - 1 - y[i];
per(j, i - 1, 0) {
if (y[j] == -1 || y[j] < y[j + 1]) break;
above = m - 1 - y[j];
sum += (ll) (below) * above * 4;
}
}
double ans = 0;
rep(i, 0, n - 1) ans += m - (y[i] >= 0);
ans = ans * ans;
ans = sum / ans;
printf("%.4f
", ans);
return 0;
}
T3 舞蹈课（dancinglessons）
也是个小傻逼题
用链表维护相邻关系
把相邻的BG扔进小根堆，每次查出最小的相邻把他们拎出来
然后剩旁边两边合并，如果也是BG那个也加入
结果a,b有一个地方打错GG了./气死
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int)a.size()
#define pb push_back
#define mp make_pair
#define ll long long
#define fi first
#define se second
const int iinf = 0x3f3f3f3f ;
const int N = 200010 ;
int n, cnt ;
string s ;
int a[N], b[N], suf[N], pre[N] ;
struct group {
int x, y, del ;
} ;
bool operator < (group a, group b) {
if (a.del != b.del) return a.del > b.del ;
else return a.x > b.x ;
}
priority_queue <group> q ;
vector <pair<int, int> > ans ;
signed main() {
// freopen("dancinglessons.in", "r", stdin) ;
// freopen("dancinglessons.out", "w", stdout) ;
scanf("%d", &n) ;
cin >> s ;
rep(i, 1, n) a[i] = (s[i - 1] == 'B') ? 1 : 0 ;
rep(i, 1, n) cnt += a[i] ;
cnt = min(cnt, n - cnt) ;
rep(i, 1, n) scanf("%d", &b[i]) ;
rep(i, 1, n - 1) suf[i] = i + 1 ;
rep(i, 2, n) pre[i] = i - 1 ;
suf[n] = pre[1] = 0 ; b[0] = iinf ;
rep(i, 1, n - 1) if (a[i] != a[i + 1]) q.push((group) {i, i + 1, abs(b[i] - b[i + 1])}) ;
rep(i, 1, cnt) {
while (!q.empty()) {
group c = q.top() ; q.pop() ;
if (b[c.x] == iinf || b[c.y] == iinf) continue ;
ans.pb(mp(c.x, c.y)) ;
suf[pre[c.x]] = suf[c.y] ;
pre[suf[c.y]] = pre[c.x] ;
int u = pre[c.x], v = suf[c.y] ;
if (a[u] != a[v]) q.push((group) {u, v, abs(b[u] - b[v])}) ;
pre[c.x] = suf[c.x] = -1 ;
pre[c.y] = suf[c.y] = -1 ;
b[c.x] = b[c.y] = iinf ;
}
}
printf("%d
", siz(ans)) ;
rep(i, 0, siz(ans) - 1) printf("%d %d
", ans[i].fi, ans[i].se) ;
return 0 ;
}