N個城市,標號從0到N-1,M條道路,第K條道路(K從0開始)的長度爲2^K,求編號爲0的城市到其他城市的最短距離
輸入描述:
第一行兩個正整數N(2<=N<=100)M(M<=500),表示有N個城市,M條道路
接下來M行兩個整數,表示相連的兩個城市的編號
輸出描述:
N-1行,表示0號城市到其他城市的最短路,如果無法到達,輸出-1,數值太大的以MOD 100000 的結果輸出。
示例1
輸入
4 4
1 2
2 3
1 3
0 1
輸出
8
9
11
解題思路:
高精度整數+dijkstra算法
看到網上還有一種看作最小生成樹問題來求解的,暫時還沒有仔細看,等有機會了再來補上這一種思路。
AC代碼:
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
#define M 100000
struct bigint {
int num[1001];
int cnt;
void init() {
for (int i = 0; i < 1001; i++) {
num[i] = 0;
}
cnt = 0;
}
void set(int x) {
init();
while (x > 0) {
num[cnt++] = x % 10;
x /= 10;
}
}
bool operator < (const bigint &a) const {
int size = max(cnt, a.cnt);
for (int i = size - 1; i >= 0; i--) {
if (num[i] < a.num[i])return true;
else if (num[i] > a.num[i]) return false;
}
return false;
}
};
struct E {
int to;
bigint cost;
};
bigint add(bigint a, bigint b) {
bigint res;
res.init();
int carry = 0;
for (int i = 0; i < a.cnt || i<b.cnt; i++) {
int tmp = a.num[i] + b.num[i] + carry;
res.num[res.cnt++] = tmp % 10;
carry = tmp / 10;
}
while (carry > 0) {
res.num[res.cnt++] = carry % 10;
carry /= 10;
}
return res;
}
bigint mul(bigint a, int x) {
bigint res;
res.init();
int carry = 0;
for (int i = 0; i < a.cnt; i++) {
int tmp = a.num[i] * x + carry;
res.num[res.cnt++] = tmp % 10;
carry = tmp / 10;
}
while (carry > 0) {
res.num[res.cnt++] = carry % 10;
carry /= 10;
}
return res;
}
int mod(bigint a, int m) {
int carry = 0;
for (int i = a.cnt - 1; i >= 0; i--) {
int t = (a.num[i] + carry * 10) / m;
int r = (a.num[i] + carry * 10) % m;
carry = r;
}
return carry;
}
void output(bigint a) {
int x = mod(a, M);
cout << x << endl;
}
vector<E> edge[101];
int n, m;
//對於距離數組d[],我們用0來表示無窮
bigint d[101]; bool mark[101];
void dijkstra(int s) {
for (int i = 0; i < 101; i++) { d[i].set(0); mark[i] = false; }
d[s].set(0); mark[s] = true; int newnode = s;
for (int i = 1; i < n; i++) {//循環n-1次
for (int j = 0; j < edge[newnode].size(); j++) {
int u = edge[newnode][j].to;
bigint c = edge[newnode][j].cost;
if (mark[u])continue;
if (d[u].cnt == 0 || add(d[newnode], c) < d[u]) {
d[u] = add(d[newnode], c);
}
}
bigint now; now.set(0);
for (int i = 0; i < n; i++) {
if (mark[i] || d[i].cnt==0)continue;
if (now.cnt == 0 || d[i] < now) {
now = d[i]; newnode = i;
}
}
mark[newnode] = true;
}
}
int main() {
scanf("%d%d", &n, &m);
bigint a;
a.set(1); int u, v;
while (m--) {
scanf("%d%d", &u, &v);
E tmp;
tmp.to = v; tmp.cost = a;
edge[u].push_back(tmp);
tmp.to = u;
edge[v].push_back(tmp);//雙向邊
a = mul(a, 2);
}
dijkstra(0);
for (int i = 1; i < n; i++) {
output(d[i]);
}
return 0;
}