LA 6476 Outpost Navigation （DFS+剪枝）

题目链接

Solution

　　DFS+剪枝

　　对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走.

记录经过每个点的最大弹药数,对dfs进行剪枝.

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
map<string, int> pos;
struct Edge {
    int v, c, next;
} edge[];
int head[], cnt;
int vis[], amm[], aim[],sum[];
int cs, n, m, ans;
string s, t;

void addedge (int u, int v, int c) {
    edge[++cnt].v = v, edge[cnt].c = c;
    edge[cnt].next = head[u];
    head[u] = cnt;
}
void dfs (int x, int s, int k) {
    if(k>=ans) return;
    if (aim[x]) {
        ans = min (ans, k);
        return ;
    }
    if (!vis[x]) s += amm[x], vis[x] = ;
    if(vis[x]&&sum[x]>=s) return;
    sum[x]=max(sum[x],s);
    for (int i = head[x]; i; i = edge[i].next) {
        int v = edge[i].v, c = edge[i].c;
        if (s >= c)  dfs (v, s - c, k + c);
    }
    vis[x] = ;
}
int main() {
    scanf ("%d", &cs);
    while (cs--) {
        pos.clear();
        memset (head, , sizeof head);
        memset(sum,,sizeof sum);
        memset(vis,,sizeof vis);
        cnt = , ans = 0x7fffffff;
        scanf ("%d %d", &n, &m);
        for (int i = ; i <= n; i++) {
            cin >> s;
            pos[s] = i;
            cin >> amm[i] >> s;
            if (s[] == 'y') aim[i] = ;
            else aim[i] = ;
        }
        for (int i = , x; i <= m; i++) {
            cin >> s >> t >> x;
            int u = pos[s], v = pos[t];
            addedge (u, v, x);
            addedge (v, u, x);
        }
        dfs (, , );
        if (ans != 0x7fffffff) printf ("%d\n", ans);
        else puts ("No safe path");
    }
}