#4：初学者之身——5

Joyoi Mobile Service，第一维：阶段，第几时刻；第二第三维：不安定因素，两个服务员的位置；省略第四维：肯定在p[i-1]。

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define inf 0x3f3f3f3f

int n, m;
int c[210][210], p[1010];
int f[2][210][210];

int main() {
    cin >> n >> m;
    rep(i, 1, n)
        rep(j, 1, n)
            cin >> c[i][j];
    rep(i, 1, m) cin >> p[i];
    p[0] = 3;

    memset(f, 0x3f, sizeof(f));
    f[0][1][2] = 0;

    rep(i, 1, m)
        rep(x, 1, n)
            rep(y, 1, n) {
                if (x != p[i-1] && y != p[i-1] && f[i-1&1][x][y] != inf) {
                    int z = p[i-1];
                    if (x != p[i] && y != p[i])    f[i&1][x][y] = min(f[i&1][x][y], f[i-1&1][x][y] + c[z][p[i]]);
                    if (y != p[i] && z != p[i]) f[i&1][y][z] = min(f[i&1][y][z], f[i-1&1][x][y] + c[x][p[i]]);
                    if (x != p[i] && z != p[i])    f[i&1][x][z] = min(f[i&1][x][z], f[i-1&1][x][y] + c[y][p[i]]);
                    f[i-1&1][x][y] = inf;
                }
            }
            
    int ans = inf;
    rep(x, 1, n)
        rep(y, 1, n)
            ans = min(ans, f[m&1][x][y]);
    cout << ans << endl;

    return 0;
}

洛谷1006，第一维：阶段，走到第几步；第二第三维：不安定因素，结尾位置；省略四维五维：有x坐标有步长可直接求y坐标。另外，一个格子只能走一次等价于走过一次还能走只不过值变成了0，会导致不如其他路线优秀。

#include <bits/stdc++.h>
using namespace std;
#define maxn 55
#define rep(i, a, b) for (int i = a; i <= b; i++)

int m, n;
int a[maxn][maxn], f[2][maxn][maxn];

int main() {
    cin >> m >> n;
    rep(i, 1, m) rep(j, 1, n)    cin >> a[i][j];

    rep(i, 0, n+m-3) {
        int st = max(i+2-n, 1);
        int ed = min(i+1, m);
        rep(j, st, ed)
            rep(k, st, ed) {
                if (j == k) {
                    f[i+1&1][j][k] = max(f[i+1&1][j][k], f[i&1][j][k] + a[j][i+3-j]);//right
                    f[i+1&1][j+1][k+1] = max(f[i+1&1][j+1][k+1], f[i&1][j][k] + a[j+1][i+2-j]);//down
                } else {
                    f[i+1&1][j][k] = max(f[i+1&1][j][k], f[i&1][j][k] + a[j][i+3-j] + a[k][i+3-k]);//right
                    f[i+1&1][j+1][k+1] = max(f[i+1&1][j+1][k+1], f[i&1][j][k] + a[j+1][i+2-j] + a[k+1][i+2-k]);//down    
                }
                if (j + 1 == k)    f[i+1&1][j+1][k] = max(f[i+1&1][j+1][k], f[i&1][j][k] + a[k][i+3-k]);//down&right
                else    f[i+1&1][j+1][k] = max(f[i+1&1][j+1][k], f[i&1][j][k] + a[k][i+3-k] + a[j+1][i+2-j]);//down&right
                if (k + 1 == j)    f[i+1&1][j][k+1] = max(f[i+1&1][j][k+1], f[i&1][j][k] + a[j][i+3-j]);//right&down
                else    f[i+1&1][j][k+1] = max(f[i+1&1][j][k+1], f[i&1][j][k] + a[j][i+3-j] + a[k+1][i+2-k]);//right&down

                f[i&1][j][k] = 0;
            }
    }

    cout << f[n+m-2 & 1][m][m];
    return 0;
}

CH Cookies，不能保证子结构后效性时说不定有排序贪心等性质，之后即可dp。下一步重头戏dp过程不会总结qwq

#include <bits/stdc++.h>
using namespace std;
#define R(a) scanf("%d", &a)
#define rep(i, a, b) for (int i = a; i <= b; i++)

int n, m;
int g[31], c[31], s[31];
int f[31][5001], a[31][5001], b[31][5001], ans[31];

inline bool cmp(int a, int b) {
    return g[a] > g[b];
}

void print(int n, int m) {
    if (!n)    return;
    print(a[n][m], b[n][m]);
    if (a[n][m] == n) {
        rep(i, 1, n)    ans[c[i]]++;
    } else {
        rep(i, a[n][m]+1, n)    ans[c[i]] = 1;
    }
}

int main() {
    R(n), R(m);
    rep(i, 1, n)    R(g[i]), c[i] = i;
    sort(c+1, c+1+n, cmp);

    memset(f, 0x3f, sizeof(f));
    f[0][0] = 0;

    rep(i, 1, n) {
        s[i] = s[i-1] + g[c[i]];
        rep(j, i, m) {
            f[i][j] = f[i][j-i];
            a[i][j] = i;
            b[i][j] = j-i;
            rep(k, 0, i-1) {
                if (f[i][j] > f[k][j - (i-k)] + k * (s[i] - s[k])) {
                    f[i][j] = f[k][j - (i-k)] + k * (s[i] - s[k]);
                    a[i][j] = k;
                    b[i][j] = j - (i-k);
                }
            }
        }
    }

    printf("%d
", f[n][m]);
    print(n, m);
    rep(i, 1, n)    printf("%d ", ans[i]);

    return 0;
}

Joyoi数字组合，初学者背包裸题。

#include <cstdio>

int n, m, v[101];
int f[10001];

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &v[i]);
    
    f[0] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = m; j >= v[i]; j--) {
            f[j] += f[j-v[i]];
        }

    printf("%d", f[m]);
    return 0;
}

BZOJ2914，有一点点像背包？就是得把题目的本质规律弄清楚就会了吧。

#include <cstdio>
#include <cctype>
#define ri readint()
#define gc getchar()
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define W(a) printf("%d
", a);
#define maxn 10005

int n, h[maxn];
int ans;
int re[maxn], p, t;
bool f[maxn];

inline int readint() {
    int x = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x * 10 + c - 48;
    return x * s;
}

int main() {
    n = ri;
    rep(i, 1, n)    h[i] = ri;
    f[h[n]+1] = true;

    rep(i, 0, h[n]+1) {
        //height i is existed but not stable!
        if (f[i] && i != h[t+1]) {
            ans = h[t];
            break;
        }
        if (i == h[t+1]) {
            if (!f[i]) {//add this as a base
                re[++p] = h[t+1];
                f[i] = true;
            }
            t++;
        }
        if (f[i]) {
            rep(j, 1, p)
                f[re[j] + i] = true;
        }
    }

    W(ans);
    rep(i, 1, p)    W(re[i]);
    return 0;
}