ZOJ 1298 Dijkstra

计算多米诺骨牌推倒推倒的时间。

这里先用Dijkstra计算单源最短路径，然后在其他不在该路径中的边所需要推倒的时间，time[i] + time[j] + edge[i][j]/2即可。这里要枚举

#include <cstdio>#include <iostream>#include <cstring>using namespace std;#define NLEN 510#define INF 1000000int edge[NLEN][NLEN];int _time[NLEN];int s[NLEN];void dijkstra(int n, int m) {memset(s, 0, sizeof(s));memset(_time, 0, sizeof(_time));for(int i = 0; i < n; i++) {_time[i] = edge[0][i];}s[0] = 1;_time[0] = 0;for(int i = 1; i < n; i++) {int min = INF;int imin = 0;for(int j = 0; j < n; j++) {if(!s[j] && _time[j] < min) {imin = j;min = _time[j];}}s[imin] = 1;for(int j = 0; j < n; j++) {if(!s[j] && edge[imin][j] < INF && _time[j] > _time[imin] + edge[j][imin]) {_time[j] = _time[imin] + edge[j][imin];}}}double max_time1 = -INF;int pos;for(int i = 0; i < n; i++) {if(_time[i] > max_time1) {max_time1 = _time[i];pos = i;}}double max_time2 = -INF;int pos1, pos2;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {if((_time[i]+_time[j]+edge[i][j])/2.0 > max_time2 && edge[i][j] < INF) {max_time2 = (_time[i] + _time[j] + edge[i][j])/2.0;pos1 = i;pos2 = j;}}}if(max_time1 >= max_time2) {printf("The last domino falls after %.1lf seconds, at key domino %d.\n\n", max_time1, pos+1);}else {printf("The last domino falls after %.1lf seconds, between key dominoes %d and %d.\n\n", max_time2, pos1+1, pos2+1);}}int main() {int n, m;int cse = 1;while(scanf("%d%d", &n, &m), n!=0) {for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {edge[i][j] = INF;}}for(int i = 0; i < m; i++) {int x, y, t;scanf("%d%d%d", &x, &y, &t);x--;y--;edge[x][y] = edge[y][x] = t;}printf("System #%d\n", cse++);dijkstra(n, m);}return 0;}