这题就是Floyd变形，只要理解了Floyd算法，题目还是不难的。不过我还是WA了一次，原理是在运行Floyd的时候，误加了两个判断，以为当i，j未被标记的时候，i到j的路径不必更新，其实这是错的，去掉以后就过了。

/*  * hdu3631/win.cpp  * Created on: 2011-9-6  * Author    : ben */ #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           using namespace std; const int SIZE = 310; const int MAX = 0x7fffffff; int map[SIZE][SIZE]; bool marked[SIZE]; int N, M; void init() {
    int i, j, temp; for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
                map[i][j] = MAX;         }         map[i][i] = 0;     } for (int k = 0; k < M; k++) {
            scanf("%d%d%d", &i, &j, &temp); if (map[i][j] > temp) {
                map[i][j] = temp;         }     } } void Floyd(int k) {
    int i, j; for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
    if (map[i][k] < map[i][j] - map[k][j]) {
                    map[i][j] = map[i][k] + map[k][j];             }         }     } } void work() {
    int T = 0, op, a, b, Q; while (scanf("%d%d%d", &N, &M, &Q) == 3) {
    if (N == 0 && M == 0 && Q == 0) {
    break;         } if (T > 0) {
                putchar('\n');         }         printf("Case %d:\n", ++T);         init();         memset(marked, false, sizeof(marked)); for (int i = 0; i < Q; i++) {
                scanf("%d", &op); if (op == 0) {
                    scanf("%d", &a); if (marked[a]) {
                        printf("ERROR! At point %d\n", a);                 } else {
                        marked[a] = true;                     Floyd(a);                 }             } else {
                    scanf("%d%d", &a, &b); if (marked[a] && marked[b]) {
    if (map[a][b] == 0x7fffffff) {
                            puts("No such path");                     } else {
                            printf("%d\n", map[a][b]);                     }                 } else {
                        printf("ERROR! At path %d to %d\n", a, b);                 }             }         }     } } int main() {
    #ifndef ONLINE_JUDGE     freopen("data.in", "r", stdin); #endif     work(); return 0; }