洛谷P1941 飞扬的小鸟(背包 dp)

题意

题目链接

sol

很显然的dp，设\(f[i][j]\)表示第\(i\)个位置，高度为\(j\)的最小步数

向上转移的时候是完全背包

向下转移判断一下就可以

#include<bits/stdc++.h>
#define fin(x) {freopen(x, "r", stdin);}
#define chmin(a, b) (a = (a < b ? a : b))
//#define int long long 
using namespace std;
const int maxn = 10001, mod = 19997, inf = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = (x * 10 + c - '0') % mod, c = getchar();
    return x * f;
}
int n, m, k;
int x[maxn], y[maxn], l[maxn], h[maxn], f[2][1001], mx, vis[maxn];//i:pos  j:height
signed main() {
    n = read(); m = read(); k = read();
    l[0] = 1; h[0] = m;
    for(int i = 1; i <= n; i++) x[i - 1] = read(), y[i - 1] = read(), l[i] = 1, h[i] = m;
    for(int i = 1; i <= k; i++) {
        int p = read(); l[p] = read() + 1; h[p] = read() - 1;
        vis[p] = 1;
    }
    for(int i = 0; i <= m; i++) f[0][i] = 0; int o = 0;
    for(int i = 0, cnt = 0; i < n; i++, o ^= 1) {
        memset(f[o ^ 1], 0x3f, sizeof(f[o ^ 1]));
        if(vis[i]) cnt++;
        for(int j = 0; j <= m; j++) {
            
            int nxt = j + x[i] > m ? m : j + x[i];
            chmin(f[o ^ 1][nxt], f[o][j] + 1);
            chmin(f[o ^ 1][nxt], f[o ^ 1][j] + 1);
            
            if(j - y[i] >= l[i + 1]) chmin(f[o ^ 1][j - y[i]], f[o][j]);
        
        }
        for(int j = 0; j < l[i + 1]; j++) f[o ^ 1][j] = inf;
        for(int j = h[i + 1] + 1; j <= m; j++) f[o ^ 1][j] = inf;
        for(int j = 0; j <= m; j++) 
            if(f[o][j] < inf && vis[i]) 
                mx = cnt;
    } 
    int ans = inf;
    for(int i = l[n]; i <= h[n]; i++) 
        ans = min(ans, f[o][i]);
    if(ans < inf) printf("1\n%d", ans);
    else printf("0\n%d", mx);
    return 0;
}