P1831 杠杆数
是否说出正解： ×
显然是个数位DP，但是感觉无从下手，事实上可以做18次数位dp，每次枚举支点，就可以算出每个数的力矩，取出力矩为0的作为贡献即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; LL read(){LL x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K,now; int str[20]; LL dp[20][3000]; LL dfs(int pos,int val,int limit){ if(!pos) return val == 0; if(val < 0) return 0; if(!limit && ~dp[pos][val]) return dp[pos][val]; LL ans = 0; int up = limit?str[pos]:9; for(int i = 0 ; i <= up; i ++){ ans += dfs(pos - 1,val + i * (pos - now),limit && (i == up)); } if(!limit) dp[pos][val] = ans; return ans; } LL solve(LL x){ if(!x) return 1; int l = 0; LL m = x; while(m){ l++; str[l] = m % 10; m /= 10; } LL ans = 0; for(int i = 1; i <= l; i ++){ Mem(dp,-1); now = i; ans += dfs(l,0,1); } return ans - (l - 1); } int main(){ LL l = read(),r = read(); Prl(solve(r) - solve(l - 1)); return 0; }
P2939 [USACO09FEB]改造路Revamping Trails
是否说出正解：√
很原题了 最短路维护dis[i][j]表示到i点已经经过了j条高速公路的最短路， Dijkstra直接维护即可
P2198 杀蚂蚁
是否说出正解：√
似乎dp[x][y]表示经过了x个放射塔，y个放射塔的最小值即可，干扰塔数量用n - x - y表示
dp[x][y] = min(dp[x - 1][y] + (x - 1) * g * (t + y * b),dp[x][y - 1] + x * g * (t + (y - 1) * b,dp[x][y] + (x + r) * g * (t + y * b)
P4766 [CERC2014]Outer space invaders
是否说出正解： ×
离散化之后区间dp
#include<bits/stdc++.h> using namespace std; const int maxn = 6010; const int INF = 0x3f3f3f3f; int N; struct Node{ int l,r,d; Node(int l = 0,int r = 0,int d = 0):l(l),r(r),d(d){} }node[maxn]; int Hash[maxn]; int dp[maxn][maxn]; int main(){ int T; scanf("%d",&T); while(T--){ scanf("%d",&N); int cnt = 0; for(int i = 1; i <= N; i ++){ scanf("%d%d%d",&node[i].l,&node[i].r,&node[i].d); Hash[++cnt] = node[i].l; Hash[++cnt] = node[i].r; } sort(Hash + 1,Hash + 1 + cnt); cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash - 1; for(int i = 1; i <= N ; i ++){ node[i].l = lower_bound(Hash + 1,Hash + 1 + cnt,node[i].l) - Hash; node[i].r = lower_bound(Hash + 1,Hash + 1 + cnt,node[i].r) - Hash; } for(int len = 1; len <= cnt; len ++){ for(int l = 1; l + len - 1 <= cnt; l ++){ int r = l + len - 1; int id = 0; for(int i = 1; i <= N ; i ++){ if(node[i].l < l || node[i].r > r) continue; if(!id || node[id].d < node[i].d) id = i; } if(!id){ dp[l][r] = 0; continue; } dp[l][r] = INF; for(int k = node[id].l; k <= node[id].r; k ++){ dp[l][r] = min(dp[l][r],dp[l][k - 1] +dp[k + 1][r] + node[id].d); } } } printf("%d\n",dp[1][cnt]); } return 0; }