UVa1078 Steam Roller——拆点+最短路

题目链接

思路

把每个点拆成(5)个点(id(x,y),id(x,y)+n,id(x,y)+2*n,id(x,y)+3*n,id(x,y)+4*n)，分别表示到这个点时的方向为上，右，下，左和静止点
非静止点的决策如下：
1.走到对应的同方向的非静止点，代价为(w)
2.走到对应的同方向的静止点，代价为(2w)
静止点的决策如下：
1.走到四联通的对应方向的非静止点，代价为(2w)
2.走到四联通的静止点，代价为(2w)
直接建图跑最短路就行了，源点为((r1,c1))对应的静止点，汇点为((r2,c2))对应的静止点
代码：

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define DEF 0x8f8f8f8f
#define DDEF 0x8f8f8f8f8f8f8f8fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline

#define N 50010
const int dir[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};

struct Edge {
  int next, to, w;
}e[5000000];

int R, C, r1, c1, r2, c2, n, S, T;
int head[N+5], eid;
int d[N+5], pre[N+5];
int W[105][105][4];
bool inq[N+5];

void addEdge(int u, int v, int w) {
  e[++eid].next = head[u];
  e[eid].to = v;
  e[eid].w = w;
  head[u] = eid;
}

int id(int x, int y) {
  return (x-1)*C+y;
}

void add(int x, int y) {
  for(int t = 0; t < 4; ++t) {
    int nx = x+dir[t][0], ny = y+dir[t][1];
    if(nx < 1 || nx > R || ny < 1 || ny > C) continue;
    int w = W[x][y][t];
    if(!w) continue;
    addEdge(id(x, y)+t*n, id(nx, ny)+t*n, w);
    addEdge(id(x, y)+t*n, id(nx, ny)+4*n, 2*w);
    addEdge(id(x, y)+4*n, id(nx, ny)+t*n, 2*w);
    addEdge(id(x, y)+4*n, id(nx, ny)+4*n, 2*w);
  }
}

void spfa() {
  memset(d, 0x3f, sizeof d);
  d[S] = 0;
  static queue<int> q;
  q.push(S);
  pre[S] = 0;
  inq[S] = 1;
  while(!q.empty()) {
    int u = q.front(); q.pop();
    inq[u] = 0;
    for(int i = head[u]; i; i = e[i].next) {
      int v = e[i].to, w = e[i].w;
      if(d[v] > d[u]+w) {
        d[v] = d[u]+w;
        pre[v] = u;
        if(!inq[v]) inq[v] = 1, q.push(v);
      }
    }
  }
}

void clear() {
  memset(head, 0, sizeof head);
  memset(W, 0, sizeof W);
  eid = 0;
}

int main() {
  int kase = 0;
  while(~scanf("%d%d%d%d%d%d", &R, &C, &r1, &c1, &r2, &c2) && R && C && r1 && c1 && r2 && c2) {
    n = R*C;
    clear();
    for(int i = 1, t; i < R; ++i) {
      for(int j = 1; j < C; ++j) {
        scanf("%d", &t);
        W[i][j][1] = W[i][j+1][3] = t;
      }
      for(int j = 1; j <= C; ++j) {
        scanf("%d", &t);
        W[i][j][2] = W[i+1][j][0] = t;
      }
    }
    for(int j = 1, t; j < C; ++j) {
      scanf("%d", &t);
      W[R][j][1] = W[R][j+1][3] = t;
    }
    for(int i = 1; i <= R; ++i)
      for(int j = 1; j <= C; ++j) add(i, j);
    S = id(r1, c1)+4*n, T = id(r2, c2)+4*n;
    spfa();
    printf("Case %d: ", ++kase);
    if(d[T] == INF) printf("Impossible
");
    else printf("%d
", d[T]);
  }
  return 0;
}