ZOJ-1655 Transport Goods 最短路

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <iomanip>
using namespace std;

/*
一个网络的运送问题，简化之后就是一个最短路问题
需要计算从N点出发到各个点的最大保有率
*/

const int MaxN = 105;
const int NONE = 2;

int N, M, trans[MaxN];
double Map[MaxN][MaxN];
bool vis[MaxN];
double rate[MaxN];

void spfa(int N) {
memset(vis, 0, sizeof (vis));
// 初始化到N点的保有率为无意义态，这个无意义态必须要求和可能出现的状态不相冲突
for (int i = 1; i <= N; ++i) {
rate[i] = NONE;
}
rate
= 1.0;
queue<int>q;
q.push(N);
vis
= true;
while (!q.empty()) {
int v = q.front();
vis[v] = false;
q.pop();
for (int i = 1; i <= N; ++i) {
if (Map[v][i] != NONE) {    // 如果有边的话
if (rate[i] == NONE) {  // 如果未被更新
rate[i] = rate[v] * Map[v][i];
q.push(i);
vis[i] = true;
} else if (rate[v] * Map[v][i] > rate[i]){
rate[i] = rate[v] * Map[v][i];
if (!vis[i]) {
q.push(i);
vis[i] = true;
}
}
}
}
}
}

int main() {
while (cin >> N >> M) {
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= N; ++j) {
Map[i][j] = NONE; // 1表示两点之间没有路相连
}
}
for (int i = 1; i < N; ++i) {
cin >> trans[i]; // 保留各个节点需要运送的量为多少
}
int a, b;
double c;
for (int i = 0; i < M; ++i) {
cin >> a >> b >> c;    // 读取边的信息，题目给定的c是一个损失率
c = 1 - c; // 得到保有率
if (Map[a][b] == NONE) {
Map[a][b] = Map[b][a] = c;
} else {
Map[a][b] = Map[b][a] = max(c, Map[a][b]);
}
}
spfa(N);
double tot = 0;
for (int i = 1; i < N; ++i) {
if (rate[i] != NONE) {
tot += trans[i] * rate[i];
}
}
cout.setf(ios::fixed);
cout << setprecision(2);
cout << tot << endl;
}
return 0;
}

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <iomanip>
using namespace std;

/*
一个网络的运送问题，简化之后就是一个最短路问题
需要计算从N点出发到各个点的最大保有率
*/

const int MaxN = 105;
const int NONE = 1;

int N, M, trans[MaxN];
double Map[MaxN][MaxN];
bool vis[MaxN];
double rate[MaxN];

void spfa(int N) {
memset(vis, 0, sizeof (vis));
// 初始化到N点的保有率为无意义态，这个无意义态必须要求和可能出现的状态不相冲突
for (int i = 1; i <= MaxN; ++i) {
rate[i] = NONE;
}
rate
= log(1.0);
queue<int>q;
q.push(N);
vis
= true;
while (!q.empty()) {
int v = q.front();
vis[v] = false;
q.pop();
for (int i = 1; i <= N; ++i) {
if (Map[v][i] != NONE) {    // 如果有边的话
if (rate[i] == NONE) {  // 如果未被更新
rate[i] = rate[v] + Map[v][i];
q.push(i);
vis[i] = true;
} else if (rate[v] + Map[v][i] > rate[i]){
rate[i] = rate[v] + Map[v][i];
if (!vis[i]) {
q.push(i);
vis[i] = true;
}
}
}
}
}
}

int main() {
while (cin >> N >> M) {
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= N; ++j) {
Map[i][j] = NONE; // 1表示两点之间没有路相连
}
}
for (int i = 1; i < N; ++i) {
cin >> trans[i]; // 保留各个节点需要运送的量为多少
}
int a, b;
double c;
for (int i = 0; i < M; ++i) {
cin >> a >> b >> c;    // 读取边的信息，题目给定的c是一个损失率
c = 1 - c; // 得到保有率
if (Map[a][b] == NONE) {
Map[a][b] = Map[b][a] = log(c); // 求指数，将概率连乘变为加法
} else {
Map[a][b] = Map[b][a] = max(log(c), Map[a][b]);
}
}
spfa(N);
double tot = 0;
for (int i = 1; i < N; ++i) {
if (rate[i] != NONE) {
tot += trans[i] * exp(rate[i]);
}
}
cout.setf(ios::fixed);
cout << setprecision(2);
cout << tot << endl;
}
return 0;
}