sicily 1321.Robot

来源：互联网发布：excel预测未来数据编辑：程序博客网时间：2024/06/15 03:56

解题策略：典型的Djstra单源路径最短问题.问题的关键是将矩阵化为图的形式.

#include <iostream>#include <vector>#include <string>#include <memory.h>#include <cstring>using namespace std;struct edge{int v;//点int cost;//权值edge(int a,int b){this->v = a;this->cost = b;}};const int MAXN = 1000005;const int N = 10005;bool used[N];int minlength[N];vector<edge> map[N];int matrix[105][105];int num[105][105];int col,row;//举证的行和列int number;//转化的图中点的个数int djstra(int start,int end){memset(used,false,sizeof(used));for(int i = 1;i <= number;i++)minlength[i] = (i == start ? 0 : MAXN);used[start] = true;for(int i = 1;i <= number;i++){int temp_min = MAXN,a = start;for(int j = 1;j <= number;j++){if(!used[j] && minlength[j] < temp_min){temp_min = minlength[j];a = j;}}used[a] = true;for(int i = 0;i < map[a].size();i++){int tempv = map[a][i].v;int cost = map[a][i].cost;if(!used[tempv] && minlength[tempv] > minlength[a] + cost)minlength[tempv] = minlength[a] + cost;}}return minlength[end];}int main(){    int n,m;    cin >> n;while(n--){int start_x,start_y,end_x,end_y;number = 0;cin >> col >> row;for(int i = 1;i <= col;i++)for(int j = 1;j <= row;j++){cin >> matrix[i][j];num[i][j] = number++;}//将矩阵转化成图for(int x = 1;x <= col;x++)for(int y = 1;y <= row;y++)//对点（x,y）的上下左右探索，若点存在，则当做从点(x,y)到此点有一条边，且权值为此点的matrix值{if(num[x][y - 1])//上{map[num[x][y]].push_back(edge(num[x][y - 1],matrix[x][y - 1]));}if(num[x + 1][y])//右{map[num[x][y]].push_back(edge(num[x + 1][y],matrix[x + 1][y]));}if(num[x][y + 1])//下{map[num[x][y]].push_back(edge(num[x][y + 1],matrix[x][y + 1]));}if(num[x - 1][y])//左{map[num[x][y]].push_back(edge(num[x - 1][y],matrix[x - 1][y]));}}cin >> start_x >> start_y >> end_x >> end_y;cout << djstra(num[start_x][start_y],num[end_x][end_y]) + matrix[start_x][start_y] << endl;for(int i = 0; i < number + 2;i++){map[i].clear();}}    return 0;}

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