DP训练（不想写博客的题目）

HDU 2196 Computer（树形dp经典）

题意：给出一棵树，求离每个节点最远的点的距离；

code：

#include 
#define ll long long
using namespace std;
const int maxn = 1e4 + 50;
struct node {
    int nex, dis;
    };
vector dge[maxn];
int d[maxn], f[maxn], son[maxn], maxxx[maxn];//d为x子树最大距离，son为子树最大距离经过的点；
//maxxx为次大的子树距离，f为x经过其父亲节点的最大距离；
bool vis[maxn];

//void dp(int x, int diss) {
//    vis[x] = true;
//    int maxx = 0;
//    int len = dge[x].size();
//    for(int i = 0; i < len; i++) {
//        int y = dge[x][i].nex;
//        if(vis[y]) continue;
//        int dis = dge[x][i].dis;
//        dp(y, diss + dis);
//        if (d[y] + dis > maxx) {
//            maxx = d[y] + dis;
//            pre[y] = x;
//            }
//        }
//    d[x] = maxx;
//    int anss = diss;
//    for (int i = 0; i < len; i++) {
//        int y = dge[x][i].nex;
//        if(vis[y])
//            continue;
//        if(pre[y] != x)
//            anss = max(anss, d[y]);
//        }
//    maxxx[x] = anss;
//    vis[x] = false;
//    return;
//    }
void dp(int x) {
    vis[x] = true;int len = dge[x].size();
    for(int i = 0; i < len; i++) {
        int y = dge[x][i].nex; int dis = dge[x][i].dis;
        if(vis[y]) continue;
        dp(y);
        if(d[y] + dis > d[x]) {
            maxxx[x] = d[x];
            d[x] = d[y] + dis;
            son[x] = y;
            }
        else if(d[y] + dis > maxxx[x]) {
            maxxx[x] = d[y] + dis;
            }
        }
    }
    
void dfs(int x) {
    vis[x] = true; int len = dge[x].size();
    for (int i = 0; i < len; i++) {
        int y = dge[x][i].nex; int dis = dge[x][i].dis;
        if(vis[y]) continue;
        if(son[x] == y)
            f[y] = max(maxxx[x],f[x])+dis;
        else
            f[y] = max(d[x],f[x])+dis;
        dfs(y);
        }
    }
    
int main() {
//    freopen("in.txt", "r", stdin);
    int n;
    while(~scanf("%d", &n)) {
        for(int i = 1; i <= n; i++) {
            dge[i].clear(); d[i] = 0; vis[i] = false;
            }
        int pos; node path;
        for(int i = 2; i <= n; i++) {
            scanf("%d%d", &pos, &path.dis);
            path.nex = i; dge[pos].push_back(path);
            path.nex = pos; dge[i].push_back(path);
            }
        int st = 1; dp(st);
        for(int i = 1; i <= n; i++) vis[i] = false; f[st] = 0;
        dfs(st);
        for (int i = 1; i <= n; i++) {
//            printf("%d\n", d[i]);
//            printf("%d\n", maxxx[i]);
//            printf("%d %d\n", i,max(f[i], d[i]));
            printf("%d\n", max(f[i], d[i]));
            }
        }
    return 0;
    }

南昌邀请赛网络赛

D Match Stick Game

#include 

using namespace std;
#define ll long long
#define ull unsigned long long
const ll INF = -1e14;
int num[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
ll st[2][10][100];
void init()
{
    for(int i = 0;i<10;i++) for(int j = 0;j<100;j++)
    {
        st[0][i][j] = -INF;
        st[1][i][j] = INF;
    }
    for(int j = 1; j <= 9; j++)
    {
        st[1][1][num[j]] = max(st[1][1][num[j]], j * 1ll);
        st[0][1][num[j]] = min(st[0][1][num[j]], j * 1ll);
    }
    for(int i = 2; i < 10; i++)
    {
        for(int j = 0; j <= 9; j++)
        {
            for(int k = 100; k >= num[j]; k--)
            {
                st[1][i][k] = max(st[1][i][k], st[1][i - 1][k - num[j]] * 10 + j);
                st[0][i][k] = min(st[0][i][k], st[0][i - 1][k - num[j]] * 10 + j);
            }
        }
    }
    st[0][1][6] = 0;
}
ll dp[105][1005];
char s[100];
int lens[105], lenn;
int main()
{
//    freopen("in.txt", "r", stdin);
    init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n; lenn = 0; int sum = 0;
        scanf("%d", &n); scanf("%s", s);
        int len = strlen(s); int cnt = 0;

        for(int i = 0; i < len; i++)
        {
            if(s[i] == '+' || s[i] == '-')
            {
                lens[lenn++] = cnt;
                cnt = 0;
            }
            else cnt++;
            if(s[i] == '-') sum += 1;
            else if(s[i] == '+') sum += 2;
            else sum += num[s[i] - '0'];
        }
        lens[lenn++] = cnt;
        for(int i = 1;i<=n;i++) for(int j = 0;j<=sum;j++) dp[i][j] = INF;
        len = lens[0];
        int totl = 2, totr = min(sum, 7*len);
        for(int j = totl; j <= totr; j++) dp[1][j] = max(dp[1][j], st[1][len][j]);
        n = lenn;
        for(int i = 2; i <= n; i++)
        {
            len = lens[i - 1];
            for(int j = totr; j >= totl; j--)
            {
                for(int k = 2 * len; k <= 7 * len; k++)
                {
                    dp[i][j + 2 + k] = max(dp[i - 1][j] + st[1][len][k], dp[i][j + 2 + k]);
                    dp[i][j + 1 + k] = max(dp[i - 1][j] - st[0][len][k], dp[i][j + 1 + k]);
                }
            }
            totr = min(sum, totr + 7 * len + 2);
            totl = totl + 2 * len + 1;
        }
        printf("%lld\n", dp[n][sum]);
    }
    return 0;
}

I Max answer

#include 
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 50;
struct Node
{
    ll val;
    int l, r;
} v[maxn];

ll s[maxn];

stack st;

int main()
{
//    freopen("in.txt", "r", stdin);
    int n; ll x; cin >> n; s[0] = 0;

    for(int i = 1; i <= n; i++)
    {
        cin >> x; s[i] = s[i - 1] + x;
        v[i].l = v[i].r = i; v[i].val = x;
    }
    
    v[0].val = -0x3f3f3f3f;
    v[n + 1].val = -0x3f3f3f3f; v[n + 1].l = v[n + 1].r = n + 1;
    
    int i = 1;
    while(i <= n + 1)
    {
        if(st.empty() || v[i].val > v[st.top()].val)
            st.push(i++);
        else
        {
            v[i].l = v[st.top()].l;
            st.pop();
        }
    }
    
    while(!st.empty()) st.pop();
    i = n + 1;
    while(i >= 1)
    {
        if(st.empty() || v[i].val > v[st.top()].val)
            st.push(i--);
        else
        {
            v[i].r = v[st.top()].r;
            st.pop();
        }
    }

    ll ans = -0x3f3f3f3f;
    for(int i = 1; i <= n; i++)
    {
//        cout << v[i].l << " " << v[i].r << endl;
        ll tem = v[i].val * (s[v[i].r] - s[v[i].l - 1]);
        ans = max(ans, tem);
    }
    ll sum = 0;
    ll minn = 0x3f3f3f3f;
    for(int i = 1; i <= n; i++)
    {
        if(sum + v[i].val < 0)
        {
            sum += v[i].val;
            minn = min(minn, v[i].val);
            ans = max(ans, minn * sum);
        }
        else sum = minn = 0;
    }
    cout << ans << endl;

    return 0;
}