洛谷P1941 飛揚的小鳥(背包 dp)
阿新 • • 發佈:2018-10-24
line else printf sig problem inline www 位置 pro
題意
題目鏈接
Sol
很顯然的dp,設\(f[i][j]\)表示第\(i\)個位置,高度為\(j\)的最小步數
向上轉移的時候是完全背包
向下轉移判斷一下就可以
#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; }
洛谷P1941 飛揚的小鳥(背包 dp)