Robot
http://acm.hdu.edu.cn/showproblem.php?pid=3028
Problem Description
There are n places (numbered 1,2,...,n), and at any second X, in place A, it will appear k robots. Now you have normal weapons. If at the second X and you are in place A, you will distroy those k robots. Otherwise, you have a special weapon (named O4), if you
at second X in the place A and use O4, you will destroy all the robots which appear in the place adjoining to A (OF COUSE include the place A). You move from place A to B (B adjoins to A) must use some second;
at second X in the place A and use O4, you will destroy all the robots which appear in the place adjoining to A (OF COUSE include the place A). You move from place A to B (B adjoins to A) must use some second;
Now you have T second ,and ask you two questions:
1. if you have an O4, how many robots can you destroy?(the maximum number);
2. if you don't have O4 but weapons, how many robots can you destroy?(the maximum number);
Input
First line: n, m, T. (n is the number of place, m is the number of roads between place, T is the seconds you have), n ≤ 100, m ≤ 2500, t ≤ 1000;
From 2 to m + 1 line: every line contains 3 integers, P, Q, D, indicating that moving from place P to place Q will use D seconds (the roads are undirection);
Next every line contains 3 integers, X, A, K, indicating at the X second, the place A will appear K robots;
The last line contains three "0" indicating that the case is end;
Output
The answer to the two question, separate by one space.
Sample Input
3 2 10 1 2 1 2 3 2 1 1 1 2 2 2 3 3 3 4 1 4 2 3 3 4 2 2 6 1 4 8 2 3 10 2 2 9 1 1 7 1 5 3 2 2 8 1 8 0 0 0
Sample Output
32 29
此代码c++能过，g++超时，详情见代码注释。
这个题目要注意一点就是x秒在a点出现k机器人可能会重复出现，例如1秒在2点出现3机器人，1秒在2点出现4机器人等等，此时我们要累加起来。
超时的原因应该是STL太慢了，改成前向星g++也能AC了。
ps:如果有大神知道更好的方法，求留言指导。
#include<cstdio>
#include<set>
#include<cstring>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
int dp[1005][105][2];//dp[i][j] 第i秒在第j点时的最大值
int times[105][1005];//times[i][j] i点在j秒有times[i][j]机器人出现
int alltime[105][1005];//alltime[i][j]预处理i点在j秒本身与相邻点所有的机器人
struct node
{
int x,t;
bool operator < (const node &a) const
{
return x<a.x;
}
} temp;
set<node> line[105];
int check(int x,int t)
{
//x相邻及本身在t秒的机器人数
int sumt=0;
for(set<node>::iterator it=line[x].begin(); it!=line[x].end(); ++it)
sumt+=times[it->x][t];
return sumt+times[x][t];
}
int main()
{
int n,m,t,p,q,d,x,a,k;
for(; ~scanf("%d%d%d",&n,&m,&t);)
{
int maxx[2];
maxx[0]=maxx[1]=0;
memset(dp,-1,sizeof(dp));
memset(times,0,sizeof(times));
for(int i=0; i<=n; ++i) line[i].clear();
//邻接表储存边
for(int i=0; i<m; ++i)
{
scanf("%d%d%d",&p,&q,&d);
temp.t=d;
temp.x=q;
line[p].insert(temp);
temp.x=p;
line[q].insert(temp);
}
for(;scanf("%d%d%d",&x,&a,&k);)
{
if(x+a+k==0) break;
times[a][x]+=k;
}
for(int i=1; i<=n; ++i)
for(int j=1; j<=t; ++j)
alltime[i][j]=check(i,j);
for(int i=1; i<=n; ++i)
{
dp[1][i][0]=times[i][1];
dp[1][i][1]=alltime[i][1];
}
for(int i=1; i<=t; ++i)
for(int j=1; j<=n; ++j)
{
if(dp[i][j][0]!=-1)
{
for(set<node>::iterator it=line[j].begin(); it!=line[j].end(); ++it)
{
if(i+it->t<=t)//转移到相邻点且目标时间不超范围
{
//从未使用O4的状态出发到相邻的点且在相邻点不使用O4
dp[i+it->t][it->x][0]=max(dp[i+it->t][it->x][0],dp[i][j][0]+times[it->x][i+it->t]);
//从未使用O4的状态出发到相邻的点且在相邻点使用O4
dp[i+it->t][it->x][1]=max(dp[i+it->t][it->x][1],dp[i][j][0]+alltime[it->x][i+it->t]);
//从使用O4的状态出发到相邻的点且在相邻点不使用O4
dp[i+it->t][it->x][1]=max(dp[i+it->t][it->x][1],dp[i][j][1]+times[it->x][i+it->t]);
//刷新最大值
maxx[0]=max(maxx[0],dp[i+it->t][it->x][0]);
maxx[1]=max(maxx[1],dp[i+it->t][it->x][1]);
}
}
if(i+1<=t)//停留在当前点的情况
{
dp[i+1][j][0]=max(dp[i+1][j][0],dp[i][j][0]+times[j][i+1]);
dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][0]+alltime[j][i+1]);
dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][1]+times[j][i+1]);
maxx[0]=max(maxx[0],dp[i+1][j][0]);
maxx[1]=max(maxx[1],dp[i+1][j][1]);
}
}
}
printf("%d %d\n",maxx[1],maxx[0]);
}
return 0;
}