Tsinsen A1105 挖地雷

http://oj.tsinsen.com/A1105

分析：简单的Dfs，用End判断是否不能再拓展出新节点，用数组记录经过的路线，没太多好说的。添加一个虚拟节点0，假定它和所有节点相连可以减少在main函数里写循环的代码量

代码：

#include "bits/stdc++.h"
using namespace std;

int Connected[15][15], G[15];
int TRes[15], Res[15], Visit[15];
int n, t, res;

void Dfs(int from, int tot, int num) {
    bool End = true;
    for (int i = 1; i <= n; ++i)
        if (!Visit[i] && Connected[from][i]) {
            Visit[i] = true; TRes[num] = i;
            Dfs(i, tot + G[i], num + 1);
            End = Visit[i] = false; TRes[num] = 0;
        }
    if (End && tot > res) {
        res = tot;
        for (int i = 1; i <= num; ++i)
            Res[i] = TRes[i];
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &G[i]);
        Connected[0][i] = 1;
    }
    for (int i = 1; i <= n - 1; ++i)
        for (int k = i + 1; k <= n; ++k) {
            scanf("%d", &t);
            Connected[i][k] = Connected[k][i] = t;
        }
    Dfs(0, G[0], 1);
    for (t = 1; Res[t] != 0; ++t)
        printf("%s%d", (t > 1) ? "-" : "", Res[t]);
    printf("\nMAX=%d\n", res);
    return 0;
}