/************************************************
* Author :Running_Time
* Created Time :2015/9/23 星期三 00:19:33
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N];
int main(void) {
int n; scanf ("%d", &n);
for (int i=1; i<=n; ++i) {
scanf ("%d", &a[i]);
}
int l = 1;
int ans = 1; int pre = a[1];
for (int i=2; i<=n; ++i) {
if (a[i] >= pre) {
pre = a[i]; ans = max (ans, i - l + 1);
}
else {
pre = a[i]; l = i;
}
}
printf ("%d
", ans);
return 0;
}
题意：每个朋友有他的金钱和友好程度，从朋友中选取一些人，问在贫富差距小于d的最大友好程度和为多少
分析：先按照m金钱从小到大排序，枚举每个起点看以这个人为最穷的人能得到的最大友好程度多少，然后是第二穷的人。。。。。复杂度O (nlogn)，尺取法也叫two points ？
/************************************************
* Author :Running_Time
* Created Time :2015/9/23 星期三 00:19:38
* File Name :B.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
struct Friend {
int m, s;
bool operator < (const Friend &r) const {
return m < r.m;
}
}f[N];
int main(void) {
int n, d, mn = INF, mx = -1;
ll sum = 0;
scanf ("%d%d", &n, &d);
for (int i=1; i<=n; ++i) {
scanf ("%d%d", &f[i].m, &f[i].s);
if (f[i].m > mx) mx = f[i].m;
if (f[i].m < mn) mn = f[i].m;
sum += f[i].s;
}
if (mx - mn < d) {
printf ("%I64d
", sum); return 0;
}
sort (f+1, f+1+n);
int l = 1, r = 2;
ll ans = f[1].s; sum = f[1].s;
while (true) {
while (r <= n && f[r].m - f[l].m < d) {
sum += f[r++].s;
}
ans = max (ans, sum);
if (r > n) break;
sum -= f[l++].s;
}
printf ("%I64d
", ans);
return 0;
}
题意：从根节点1出发，到底部的点的路径没有超过连续m个点有cat的个数为多少
分析：简单的深搜，记录走到当前点最大连续cat数，注意一些剪枝
/************************************************
* Author :Running_Time
* Created Time :2015/9/23 星期三 00:19:41
* File Name :C.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
vector<int> G[N];
int cat[N];
bool vis[N];
int n, m, ans;
void DFS(int u, int fa, int num) {
for (int i=0; i<G[u].size (); ++i) {
int v = G[u][i];
if (vis[v] || v == fa) continue;
if (cat[v]) {
if (num + 1 <= m) { //算上当前点，cat + 1
if (G[v].size () == 1) { //到达底部
vis[v] = true; ans++;
}
else {
vis[v] = true; //未到达底部
DFS (v, u, num + 1);
}
}
else continue; //条件不符，不往下艘
}
else {
if (num <= m) {
if (G[v].size () == 1) {
vis[v] = true; ans++;
}
else {
vis[v] = true;
DFS (v, u, 0);
}
}
else continue;
}
}
}
int main(void) {
scanf ("%d%d", &n, &m);
for (int i=1; i<=n; ++i) {
scanf ("%d", &cat[i]);
}
for (int u, v, i=1; i<n; ++i) {
scanf ("%d%d", &u, &v);
G[u].push_back (v);
G[v].push_back (u);
}
DFS (1, 0, cat[1]);
printf ("%d
", ans);
return 0;
}
状态压缩DP D - Kefa and Dishes
题意：n道菜选择m道，每道菜有一个愉悦度，如果某些菜按照先后顺序吃还能得到额外的愉悦度，问最大愉悦度为多少
分析：其实就是一个DAG问题，数据范围18应该想到状压，我不熟悉以为数据范围太大，做不了。dp[mask][i] 表示当在mask集合状态下，最后是第i道菜的最大愉悦度为多少，状态转移方程：dp[(i|(1<<l))][l] = max (dp[i][j] + a[l] + g[j][l]) 复杂度O(2n * n2).
/************************************************
* Author :Running_Time
* Created Time :2015/9/23 星期三 01:49:58
* File Name :D.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 18;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int n, m, k;
ll dp[(1<<N)+10][N];
int a[N];
int g[N][N];
int main(void) {
scanf ("%d%d%d", &n, &m, &k);
for (int i=0; i<n; ++i) {
scanf ("%d", &a[i]);
}
for (int u, v, c, i=1; i<=k; ++i) {
scanf ("%d%d%d", &u, &v, &c);
if (g[u-1][v-1] < c) g[u-1][v-1] = c;
}
int maxS = (1 << n);
for (int i=0; i<n; ++i) {
dp[(1<<i)][i] = a[i];
}
for (int i=1; i<maxS; ++i) {
for (int j=0; j<n; ++j) {
if ((i & (1 << j)) == 0) continue;
for (int l=0; l<n; ++l) {
if ((i & (1 << l)) == 0) {
if (dp[(i|(1<<l))][l] < dp[i][j] + a[l] + g[j][l]) {
dp[(i|(1<<l))][l] = dp[i][j] + a[l] + g[j][l];
}
}
}
}
}
ll ans = 0;
for (int i=1; i<maxS; ++i) {
int cnt = 0;
for (int j=0; j<n; ++j) {
if ((i & (1 << j)) != 0) cnt++;
}
if (cnt == m) {
for (int j=0; j<n; ++j) {
if ((i & (1 << j)) != 0) {
if (ans < dp[i][j]) ans = dp[i][j];
}
}
}
}
printf ("%I64d
", ans);
return 0;
}