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  • 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
    %d", ans);
    	else printf("0
    %d", mx);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9841089.html
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