uva 12222——Mountain Road
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题意:有一个狭窄的山路,两辆车不能同时相向而行,并且同向的车的间隔不能小于10秒,给定一些车的起始时间,求最后一辆车离开的最短时间。
思路:递推,dp(i,j,k)表示A端过了i辆车,B端过了j辆车,下辆车从k端走的情况,对于A端下一个车枚举i+1的车情况更新,B端则枚举第j+1辆车来更新。
code:
#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=9999999;//0x3fffffff;const int inf=-INF;const int N=1000000;const int M=2005;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define ft(i,s,n) for (int i=s;i<=n;i++)#define frt(i,s,n) for (int i=s;i>=n;i--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt rt<<1#define rrt rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);int len1,len2;int n,dp[M][M][2];struct node{ int s,t;}g[2][M];int main(){ int T; scanf("%d",&T); while (T--){ scanf("%d",&n); char c; int s,t; len1=len2=0; ft(i,1,n){ getchar(); //scanf("%c",&c); //printf("%c\n",c); scanf("%c%d%d",&c,&s,&t); if (c=='A'){ len1++; g[0][len1].s=s; g[0][len1].t=t; } else{ len2++; g[1][len2].s=s; g[1][len2].t=t; } } int ts,td; cls(dp,INF); dp[0][0][0]=dp[0][0][1]=0; ft(i,0,len1){ ft(j,0,len2){ ts=dp[i][j][1],td=0; ft(k,i+1,len1){ ts=max(ts,g[0][k].s); td=max(td,ts+g[0][k].t); dp[k][j][0]=min(dp[k][j][0],td); ts+=10;td+=10; } ts=dp[i][j][0],td=0; ft(k,j+1,len2){ ts=max(ts,g[1][k].s); td=max(td,ts+g[1][k].t); dp[i][k][1]=min(dp[i][k][1],td); ts+=10;td+=10; } } } printf("%d\n",min(dp[len1][len2][0],dp[len1][len2][1])); }}
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