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 << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[N];int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_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 << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m;typedef pair<int, int> P;P a[N];int deq[N];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 << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, k, ans;bool cat[N];vector<int> G[N];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 LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_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 << endlconst 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;}