HDU 4568 Hunter（最短路径+DP）（2013 ACM-ICPC长沙赛区全国邀请赛）

[align=left]Problem Description[/align]
　　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.

#include <cstdio>
#include <queue>
#include <utility>
#include <iostream>
#include <cstring>
using namespace std;
typedef pair<int, int> PII;

const int MAXN = 205;

int mat[MAXN][MAXN], tx[15], ty[15];
int dis[MAXN][MAXN], di[15][15];
bool ist[MAXN][MAXN];
int post[MAXN][MAXN];
int n, m, k;

#define pos(x, y) (x*MAXN+y)

int fx[4] = {-1,0,1,0};
int fy[4] = {0,1,0,-1};

void min_path(int st_x, int st_y, int now) {
priority_queue<PII> que;
que.push(make_pair(-mat[st_x][st_y], pos(st_x, st_y)));
memset(dis, 0x3f, sizeof(dis));
dis[st_x][st_y] = mat[st_x][st_y];
while(!que.empty()) {
int abc = -que.top().first, tmp = que.top().second; que.pop();
int x = tmp / MAXN, y = tmp % MAXN;
if(abc != dis[x][y]) continue;
for(int i = 0; i < 4; ++i) {
int newx = x + fx[i], newy = y + fy[i];
if(0 <= newx && newx < n && 0 <= newy && newy < m) {
if(mat[newx][newy] == -1) continue;
if(dis[newx][newy] > mat[newx][newy] + dis[x][y]) {
dis[newx][newy] = mat[newx][newy] + dis[x][y];
que.push(make_pair(-dis[newx][newy], pos(newx, newy)));
if(ist[newx][newy] && dis[newx][newy] < di[now][post[newx][newy]])
di[now][post[newx][newy]] = dis[newx][newy];
}
}
else if(dis[x][y] < di[now][k]) di[now][k] = dis[x][y];
}
}
}

int dp[14][20000];
int ans, sum;

int dfs(int u, int use) {
if(dp[u][use] < 0x3f3f3f3f) return dp[u][use];
if(use == 0) {
return dp[u][use] = di[u][k];
}
for(int i = 0; i < k; ++i) {
if(use & (1 << i))
dp[u][use] = min(dp[u][use], di[u][i] + dfs(i, use ^ (1 << i)));
}
return dp[u][use];
}

void solve() {
memset(dp, 0x3f, sizeof(dp));
ans = 0x7fff7fff;
for(int i = 0; i < k; ++i)
ans = min(ans, di[i][k] + dfs(i, (1 << i) ^ ((1 << k) - 1)));
}

void printdi() {
for(int i = 0; i < k; ++i) {
for(int j = 0; j <= k; ++j) printf("%d ", di[i][j]);
printf("\n");
}
}

int main() {
int T;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j) scanf("%d", &mat[i][j]);
scanf("%d", &k);
memset(ist, 0, sizeof(ist));
sum = 0;
for(int i = 0; i < k; ++i) {
scanf("%d%d", &tx[i], &ty[i]);
ist[tx[i]][ty[i]] = true;
post[tx[i]][ty[i]] = i;
sum += mat[tx[i]][ty[i]];
}
memset(di, 0x3f, sizeof(di));
for(int i = 0; i < k; ++i) min_path(tx[i], ty[i], i);
//printdi();
solve();
printf("%d\n", ans - sum);
}
}

View Code