HDU-4568 Hunter （Dijkstra&&状压DP）
Hunter
http://acm.hdu.edu.cn/showproblem.php?pid=4568
Problem Description
One day, a hunter named James went to a mysterious area to find the treasures. James wanted to research the area and brought all treasures that he could.
The area can be represented as a N*M rectangle. Any points of the rectangle is a number means the cost of research it,-1 means James can't cross it, James can start at any place out of the rectangle, and explore point next by next. He will move in the rectangle and bring out all treasures he can take. Of course, he will end at any border to go out of rectangle(James will research every point at anytime he cross because he can't remember whether the point are researched or not).
Now give you a map of the area, you must calculate the least cost that James bring out all treasures he can take(one point up to only one treasure).Also, if nothing James can get, please output 0.
The area can be represented as a N*M rectangle. Any points of the rectangle is a number means the cost of research it,-1 means James can't cross it, James can start at any place out of the rectangle, and explore point next by next. He will move in the rectangle and bring out all treasures he can take. Of course, he will end at any border to go out of rectangle(James will research every point at anytime he cross because he can't remember whether the point are researched or not).
Now give you a map of the area, you must calculate the least cost that James bring out all treasures he can take(one point up to only one treasure).Also, if nothing James can get, please output 0.
Input
The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case begins with a line containing 2 integers N M , (1<=N,M<=200), that represents the rectangle. Each of the following N lines contains M numbers(0~9),represent the cost of each point. Next is K(1<=K<=13),and next K lines, each line contains 2 integers x y means the position of the treasures, x means row and start from 0, y means column start from 0 too.
Output
For each test case, you should output only a number means the minimum cost.
Sample Input
2 3 3 3 2 3 5 4 3 1 4 2 1 1 1 3 3 3 2 3 5 4 3 1 4 2 2 1 1 2 2
Sample Output
8 11
先做的hdu-4571感觉和那题差不多，都需要先处理出任意两点的最短路，不过本题少了一个限制，所以dfs没有很好的终止条件的，然后就不会了...
看了题解后，发现必经点少，可以进行状态压缩储存已经过的点，就想到以前做过更相似的题（hdu-5418），才发现这都是TSP问题
为了方便，增加虚拟的开始与终止节点，不过这样会增加很大的常数
手快打错了，导致又浪费好多时间，找错时想到好多符合题目描述的点，但数据都没有出到...
真是 脑抽1秒钟，调试n小时。
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN=205;
const int INF=0x3f3f3f3f;
const int dr[4]={-1,0,1,0};
const int dc[4]={0,1,0,-1};
struct Node {
int r,c;
Node(int rr=0,int cc=0):r(rr),c(cc) {}
bool operator < (const Node& a) const {
return r<a.r||(r==a.r&&c<a.c);
}
}p[MAXN];
int n,m,k,STA,DES;
int mz[MAXN][MAXN],dis[MAXN*MAXN],cost[15][15],dp[(1<<16)+1][15];//dp[i][j]表示当前i中为1的点已探索过，且最后一个点是j
bool vis[MAXN*MAXN];
inline bool isInside(int r,int c) {
return 0<=r&&r<n&&0<=c&&c<m;
}
void Dijkstra(int sta) {
int r,c,rr,cc,tmp;
Node u;
priority_queue<Node> q;
tmp=p[sta].r*m+p[sta].c;
q.push(Node(0,tmp));//偷懒，用r代表当前点与sta的距离，c代表当前点
memset(dis,0x3f,sizeof(dis));
memset(vis,false,sizeof(vis));
dis[tmp]=0;//dis[tmp]表示sta点到tmp点的距离，不包括sta点的花费，包括tmp点的花费
while(!q.empty()) {
do {
u=q.top();
q.pop();
} while(!q.empty()&&vis[u.c]);
if(vis[u.c])
return ;
vis[u.c]=true;
dis[u.c]=-u.r;
r=u.c/m;
c=u.c%m;
for(int i=0;i<4;++i) {
rr=r+dr[i];
cc=c+dc[i];
tmp=rr*m+cc;
if(isInside(rr,cc)&&!vis[tmp]&&mz[rr][cc]!=-1)
q.push(Node(-dis[u.c]-mz[rr][cc],tmp));
}
}
}
int main() {
int T,mn,t,kk;
scanf("%d",&T);
while(T-->0) {
scanf("%d%d",&n,&m);
for(int i=0;i<n;++i)
for(int j=0;j<m;++j)//最开始m写成n，查了好久错...
scanf("%d",&mz[i][j]);
scanf("%d",&k);
STA=0;
DES=k+1;
for(int i=1;i<=k;++i)
scanf("%d%d",&p[i].r,&p[i].c);
memset(cost,0x3f,sizeof(cost));
for(int i=1;i<=k;++i) {
Dijkstra(i);
for(int j=1;j<=k;++j)
cost[i][j]=dis[p[j].r*m+p[j].c];
mn=INF;
for(int j=0;j<n;++j)
mn=min(mn,min(dis[j*m],dis[j*m+m-1]));
t=n*m-m;
for(int j=0;j<m;++j)
mn=min(mn,min(dis[j],dis[t+j]));
cost[i][DES]=mn;//出去的最小花费即该点到边界的最小花费
cost[STA][i]=mz[p[i].r][p[i].c]+mn;//此处要包含探索i的花费
}
k+=2;
kk=1<<k;
memset(dp,0x3f,sizeof(dp));
dp[1<<STA][STA]=0;
for(int i=0;i<kk;++i)
for(int j=0;j<k;++j)
if(((i>>j)&1)==1)
for(int l=0;l<k;++l)
dp[i][j]=min(dp[i][j],dp[i&(~(1<<j))][l]+cost[l][j]);
printf("%d\n",dp[kk-1][DES]==INF?0:dp[kk-1][DES]);
}
return 0;
}