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题意:题意难懂,就是一个小偷在m天内从城市1飞到城市n最小花费,输入的是每个城市飞到其他城市的航班。
分析:dp[i][j] 表示小偷第i天在城市j的最小花费。状态转移方程:dp[i][j] = min (dp[i-1][k] + cost[k][j][t%day]) t表示在t天时k飞往j的飞机的花费
收获:
代码:
/************************************************
* Author :Running_Time
* Created Time :2015-8-29 14:07:43
* File Name :UVA_590.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dp[1010][12];
int d[12][12];
int cost[12][12][32];
int main(void) {
int n, m, cas = 0;
while (scanf ("%d%d", &n, &m) == 2) {
if (n == 0 && m == 0) break;
for (int i=1; i<=n; ++i) {
for (int j=1; j<=n; ++j) {
if (i != j) {
scanf ("%d", &d[i][j]);
for (int k=0; k<d[i][j]; ++k) {
scanf ("%d", &cost[i][j][k]);
}
}
}
}
memset (dp, INF, sizeof (dp));
for (int i=2; i<=n; ++i) {
if (cost[1][i][0]) {
dp[0][i] = cost[1][i][0];
}
}
for (int i=1; i<m; ++i) {
for (int k=1; k<=n; ++k) {
for (int j=1; j<=n; ++j) {
if (j != k) {
int c = cost[j][k][i%d[j][k]];
if (c) dp[i][k] = min (dp[i][k], dp[i-1][j] + c);
}
}
}
}
int ans = dp[m-1][n];
printf("Scenario #%d\n", ++cas);
if(ans != INF){
printf("The best flight costs %d.\n\n", ans);
}else{
puts("No flight possible.\n");
}
}
return 0;
}
递推DP UVA 590 Always on the run
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原文地址:http://www.cnblogs.com/Running-Time/p/4773806.html