uva 1632 Alibaba

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题目大意：lrj306页习题9-8 想了很久，区间DP的状态意识到了，但是不知道这个怎么和deadline联系起来，最后看别人的解题报告，恍然大悟，原来如此，顺便练了一下滚动数组，还是很有收获的。

题目链接：https://vjudge.net/problem/UVA-1632

第一份代码没有使用滚动数组，第二份使用了滚动数组，第二份比第一份的速度要上不少。

#include <cstdio>#include <iostream>#include <cstring>#define min(a,b) (((a)<(b))?(a):(b))using namespace std;const int maxn = 10005 , inf = 0x3f3f3f3f;int a[maxn] , d[maxn];int dp[maxn][maxn][2];int n;void solve () {    for (int i = 1 ; i <= n ; i++)        dp[i][i][0] = dp[i][i][1] = 0;    for (int i = n-1 ; i >= 1 ; i--) {        for (int j = i+1 ; j <= n ; j++) {            dp[i][j][0] = min(dp[i+1][j][0]+a[i+1]-a[i] , dp[i+1][j][1]+a[j]-a[i]);            if (dp[i][j][0] >= d[i]) dp[i][j][0] = inf;            dp[i][j][1] = min(dp[i][j-1][1]+a[j]-a[j-1] , dp[i][j-1][0]+a[j]-a[i]);            if (dp[i][j][1] >= d[j]) dp[i][j][1] = inf;        }    }    int ans = min(dp[1][n][0] , dp[1][n][1]);    if (ans >= inf)        cout << "No solution" << endl;    else        cout << ans << endl;}int main () {    while (cin >> n) {        memset (a,0,sizeof(a));        memset (d,0,sizeof(d));        memset (dp,0,sizeof(dp));        for (int i = 1 ; i <= n ; i++)            scanf("%d%d" , &a[i] , &d[i]);        solve();    }}

#include <cstdio>#include <iostream>#include <cstring>#define min(a,b) (((a)<(b))?(a):(b))using namespace std;const int maxn = 10005 , inf = 0x3f3f3f3f;int a[maxn] , d[maxn];int dp[2][maxn][2];int n;void solve () {    dp[(n&1)^1][n][1] = dp[(n&1)^1][n][0] = 0;    for (int i = n-1 ; i >= 1 ; i--) {        int x = (i&1)^1;        for (int j = i+1 ; j <= n ; j++) {            dp[x][j][0] = min(dp[x^1][j][0]+a[i+1]-a[i] , dp[x^1][j][1]+a[j]-a[i]);            if (dp[x][j][0] >= d[i]) dp[x][j][0] = inf;            dp[x][j][1] = min(dp[x][j-1][1]+a[j]-a[j-1] , dp[x][j-1][0]+a[j]-a[i]);            if (dp[x][j][1] >= d[j]) dp[x][j][1] = inf;        }    }    int ans = min(dp[0][n][0] , dp[0][n][1]);    if (ans >= inf)        cout << "No solution" << endl;    else        cout << ans << endl;}int main () {    while (cin >> n) {        memset (a,0,sizeof(a));        memset (d,0,sizeof(d));        memset (dp,0,sizeof(dp));        for (int i = 1 ; i <= n ; i++)            scanf("%d%d" , &a[i] , &d[i]);        solve();    }}

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