Codeforces #321（div2）

A. Kefa and First Steps
题意: 求连续最长不降子序列

分析: 水题

代码:

//
//  Created by TaoSama on 2015-09-23
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a
;

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d", &n) == 1){
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
a[n + 1] = -INF;
int ans = 1, cnt = 1;
for(int i = 2; i <= n + 1; ++i){
if(a[i] >= a[i - 1]) ++cnt;
else{
ans = max(ans, cnt);
cnt = 1;
}
}
printf("%d\n", ans);
}
return 0;
}

B. Kefa and Company
题意: 给很多对身价和满意度的对 要求选择的对中任意2个对 身价差距不超过d 求最大满意度
分析: 按照身价排序 然后维护最大身价-最小身价不超过d的滑动窗口 更新最大满意度就是答案
代码:

//
//  Created by TaoSama on 2015-09-23
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
typedef pair<int, int> P;
P a
;
int deq
;

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &m) == 2) {
for(int i = 1; i <= n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
sort(a + 1, a + 1 + n);
int l = 1;
long long sum = 0, ans = 0;
for(int i = 1; i <= n; ++i) {
sum += a[i].second;
deq[i] = i;
while(i > l
&& a[deq[i]].first - a[deq[l]].first >= m) sum -= a[deq[l++]].second;
//          printf("%d\n", sum);
ans = max(ans, sum);
}
printf("%I64d\n", ans);
}
return 0;
}

C. Kefa and Park

题意: 求树上到叶子节点的路径中 连续猫个数不超过m 的不同叶子节点
分析: 搜一下树就好了
代码:

//
//  Created by TaoSama on 2015-09-23
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, k, ans;
bool cat
;
vector<int> G
;

void dfs(int u, int f, int c){
if(c > k) return;
int cnt = 0;
for(auto v : G[u]){
if(v == f) continue;
++cnt;
dfs(v, u, cat[v] ? c + 1 : 0);
}
if(cnt == 0) ++ans;
}

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &k) == 2){
for(int i = 1; i <= n; ++i) scanf("%d", cat + i);
for(int i = 1; i <= n; ++i) G[i].clear();
for(int i = 1; i < n; ++i){
int u, v; scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
ans = 0;
dfs(1, -1, cat[1]);
printf("%d\n", ans);
}
return 0;
}

D. Kefa and Dishes

题意: n个饭菜 吃m个 如果按照k种前后排列的话有额外奖励 求最大满意度
分析: 状压dp 需要知道当前谁吃了 当前谁吃了 所以dp[i][j]:= 当前状态为i 且吃了j的最大满意度
          然后dp就好了
代码:

//
//  Created by TaoSama on 2015-09-23
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, k;
long long dp[1 << 18][20];
int a[20], c[20][20];

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d%d", &n, &m, &k) == 3) {
for(int i = 0; i < n; ++i) scanf("%d", a + i);
memset(c, 0, sizeof c);
for(int i = 1; i <= k; ++i) {
int u, v, w; scanf("%d%d%d", &u, &v, &w);
--u; --v;
c[u][v] = w;
}
memset(dp, -1, sizeof dp);
for(int i = 0; i < n; ++i) dp[1 << i][i] = a[i];
long long ans = 0;
for(int i = 1; i < 1 << n; ++i) {
for(int j = 0; j < n; ++j) {
if(dp[i][j] == -1) continue;
for(int k = 0; k < n; ++k) {
if(i >> k & 1) continue;
dp[i | (1 << k)][k] = max(dp[i | (1 << k)][k], dp[i][j] + c[j][k] + a[k]);
}
if(__builtin_popcount(i) == m) ans = max(ans, dp[i][j]);
}
}
printf("%I64d\n", ans);
}
return 0;
}