用c语言实现本题

好的，以下是C语言代码实现，使用了Dijkstra算法求解最短路径和记录沿途解放城镇数和歼伤敌军数的方法：

#include <stdio.h>
#include <string.h>
#define N 210
#define INF 0x3f3f3f3f

int n, m, cnt;
int graph[N][N], dist[N], path[N], num[N], kill[N], vis[N];
char start[4], end[4], city[N][4];

int get_index(char str[]) {
    for (int i = 1; i <= n; i++) {
        if (strcmp(city[i], str) == 0)
            return i;
    }
    return 0;
}

void dijkstra(int start) {
    memset(vis, 0, sizeof(vis));
    memset(dist, INF, sizeof(dist));
    memset(num, 0, sizeof(num));
    memset(kill, 0, sizeof(kill));
    dist[start] = 0;
    num[start] = 1;
    kill[start] = graph[start][start];
    for (int i = 1; i <= n; i++) {
        int u = 0;
        for (int j = 1; j <= n; j++) {
            if (!vis[j] && (u == 0 || dist[j] < dist[u]))
                u = j;
        }
        vis[u] = 1;
        for (int v = 1; v <= n; v++) {
            if (!vis[v] && graph[u][v] != INF) {
                if (dist[u] + graph[u][v] < dist[v]) {
                    dist[v] = dist[u] + graph[u][v];
                    num[v] = num[u] + 1;
                    kill[v] = kill[u] + graph[u][v];
                    path[v] = u;
                } else if (dist[u] + graph[u][v] == dist[v]) {
                    if (num[u] + 1 > num[v]) {
                        num[v] = num[u] + 1;
                        kill[v] = kill[u] + graph[u][v];
                        path[v] = u;
                    } else if (num[u] + 1 == num[v]) {
                        if (kill[u] + graph[u][v] > kill[v]) {
                            kill[v] = kill[u] + graph[u][v];
                            path[v] = u;
                        }
                    }
                }
            }
        }
    }
}

void print_path(int end) {
    if (end == 0)
        return;
    print_path(path[end]);
    printf("->%s", city[end]);
}

int main() {
    scanf("%d%d%s%s", &n, &m, start, end);
    for (int i = 1; i < n; i++) {
        char name[4];
        int num;
        scanf("%s%d", name, &num);
        int index = get_index(name);
        strcpy(city[index], name);
        graph[1][index] = graph[index][1] = num;
    }
    for (int i = 0; i < m; i++) {
        char name1[4], name2[4];
        int len;
        scanf("%s%s%d", name1, name2, &len);
        int index1 = get_index(name1);
        int index2 = get_index(name2);
        graph[index1][index2] = graph[index2][index1] = len;
    }
    int start_index = get_index(start);
    int end_index = get_index(end);
    dijkstra(start_index);
    printf("%s", start);
    print_path(end_index);
    printf("\n%d %d %d", num[end_index], dist[end_index], kill[end_index]);
    return 0;
}

注意：在使用Dijkstra算法时，需要对每个节点初始化。在本题中，需要同时记录沿途解放城镇数和歼伤敌军数，所以需要额外开辟数组分别记录。同时需要注意在输出路径时，先输出起点，再倒序输出路径上的节点。